Meet Your Tutor
Integration improves when you diagnose the form before choosing a method. I will help you look for a direct antiderivative, substitution, identity, partial fraction or by-parts structure in that order, and verify each result by differentiation whenever possible.
Let us start with the honest bit. Integration is the chapter where a lot of Class 12 students quietly decide that maths is “not for them”. You look at a question, you have twelve formulas in your head, and not one of them seems to fit. That feeling is completely normal, and it is not a sign that you are bad at maths. It is a sign that nobody has yet shown you the map.
So here is the promise for this chapter. We are going to build integration from absolute zero — starting from one single idea (integration just undoes differentiation) — and then add one tool at a time. Substitution. Trigonometric identities. Partial fractions. By parts. The special-type integrals. Definite integrals and their properties. Every single method gets its own section, its own worked examples, and its own “how do I spot this?” signal.
Take it slowly. Read one section, do its examples with a pen in your hand, and only then move on. There is no prize for finishing fast. Integrals sits inside Unit III — Calculus, which carries 35 of the 80 marks in your Class 12 Maths paper, so the time you spend here pays back more than almost anywhere else in the syllabus.
What You’ll Learn
- Integration As The Inverse Of Differentiation
- The Standard Formula List You Must Know
- Basic Rules And Rewriting Before You Integrate
- Integration By Substitution
- Integrals Using Trigonometric Identities
- Integration By Partial Fractions
- Integration By Parts
- The Special Result For eˣ[f(x) + f′(x)]
- Special Type Integrals: The a² And x² Family
- Completing The Square For ax² + bx + c
- Integrals Of The Type (px + q) Over A Quadratic
- Integrals Involving Square Roots Of Quadratics
- Choosing The Right Method
- Definite Integrals And The Fundamental Theorem
- Definite Integrals By Substitution
- Properties Of Definite Integrals
- Putting The Properties To Work
- Common Mistakes And How To Avoid Them
Your Game Plan
- Learn the standard formula list first. Everything else in this chapter reduces to one of those forms. Write them on a card and revise them daily until they are automatic.
- Master substitution next. It is the single most-used technique, and it is what turns an ugly integral into a formula-list integral.
- Then take partial fractions and by parts. These two are the big mark-earners. Do not rush them.
- Drill the special-type integrals. Completing the square is the one skill that unlocks the entire family listed in your syllabus.
- Only then move to definite integrals — the Fundamental Theorem, then the properties. The properties feel like magic tricks until you have done ten of them; after that they feel obvious.
- Finish with the worksheet at the end. Attempt every question with the answer hidden. Reveal only after you have written something down.
Study Notes
Integration As The Inverse Of Differentiation
Imagine somebody hands you a photograph of a cake and asks you to work out the recipe. That is integration. Differentiation is baking: you start with a function and you produce its derivative. Integration is the reverse walk — you are given the derivative and asked to name the function it came from.
Formally: if d/dx [F(x)] = f(x), then we say F(x) is an anti-derivative (or primitive) of f(x), and we write ∫ f(x) dx = F(x) + C. The function f(x) is called the integrand, and dx tells you which variable you are integrating with respect to.
Now, why the + C? Because differentiation destroys information. The derivative of x² is 2x. But so is the derivative of x² + 7, and of x² − 1000. Constants vanish when you differentiate, so when you walk backwards you have no way of knowing what constant was there. We admit that honestly by writing “+ C”, an arbitrary constant. Geometrically, ∫ f(x) dx is not one curve but a whole family of parallel curves, each shifted vertically from the next.
Key Idea — the two-way street
∫ f(x) dx = F(x) + C means exactly the same thing as d/dx[F(x) + C] = f(x). These are two ways of saying one fact. This gives you a free superpower: you can always check an integration answer by differentiating it. If differentiating your answer gives back the integrand, you are right. Use this in the exam — it costs twenty seconds and catches almost every sign slip.
Example 1 — Reading the definition backwards
Find ∫ 4x³ dx.
Ask yourself: what function, when differentiated, gives 4x³? We know d/dx(x⁴) = 4x³. That is an exact match.
Answer: ∫ 4x³ dx = x⁴ + C.
Check: d/dx(x⁴ + C) = 4x³. ✓
Ask yourself: what function, when differentiated, gives 4x³? We know d/dx(x⁴) = 4x³. That is an exact match.
Answer: ∫ 4x³ dx = x⁴ + C.
Check: d/dx(x⁴ + C) = 4x³. ✓
Example 2 — Adding pieces
Find ∫ (2x + cos x) dx.
Handle the pieces separately. Something whose derivative is 2x is x². Something whose derivative is cos x is sin x.
Answer: x² + sin x + C.
Check: d/dx(x² + sin x + C) = 2x + cos x. ✓ Notice we write only one constant C at the end, not one per term — a sum of arbitrary constants is just another arbitrary constant.
Handle the pieces separately. Something whose derivative is 2x is x². Something whose derivative is cos x is sin x.
Answer: x² + sin x + C.
Check: d/dx(x² + sin x + C) = 2x + cos x. ✓ Notice we write only one constant C at the end, not one per term — a sum of arbitrary constants is just another arbitrary constant.
Good to Know — a note on scope
You may have seen older material that develops the definite integral as “the limit of a sum” (the Σ formula with n → ∞). That topic is not listed in the CBSE Class 12 Maths curriculum text for the 2026-27 session, so this chapter does not build worked examples around it. It is still lovely mathematics and worth reading once for understanding. Always confirm against the syllabus copy your own school has issued for your session.
The Standard Formula List You Must Know
This table is the foundation of the whole chapter. Every technique you will learn — substitution, by parts, partial fractions, completing the square — exists for one reason: to bend a difficult integral into one of these shapes. Do not try to memorise them by staring. Instead, read each row right to left and confirm the derivative. When you can do that for every row, you own the table.
| Integral | Result | Remember this |
|---|---|---|
| ∫ xⁿ dx | xⁿ⁺¹/(n+1) + C | valid for n ≠ −1 |
| ∫ (1/x) dx | log |x| + C | this is the missing n = −1 case |
| ∫ eˣ dx | eˣ + C | the function that is its own derivative |
| ∫ aˣ dx | aˣ/log a + C | a > 0, a ≠ 1 |
| ∫ cos x dx | sin x + C | — |
| ∫ sin x dx | − cos x + C | watch the minus sign |
| ∫ sec² x dx | tan x + C | — |
| ∫ cosec² x dx | − cot x + C | — |
| ∫ sec x tan x dx | sec x + C | — |
| ∫ cosec x cot x dx | − cosec x + C | — |
| ∫ tan x dx | log |sec x| + C | same as − log |cos x| + C |
| ∫ cot x dx | log |sin x| + C | — |
| ∫ sec x dx | log |sec x + tan x| + C | — |
| ∫ cosec x dx | log |cosec x − cot x| + C | same as log |tan(x/2)| + C |
| ∫ dx/(x² − a²) | (1/2a) log |(x − a)/(x + a)| + C | bigger term first, inside the bracket too |
| ∫ dx/(a² − x²) | (1/2a) log |(a + x)/(a − x)| + C | mirror image of the row above |
| ∫ dx/(x² + a²) | (1/a) tan⁻¹(x/a) + C | a plus sign always gives you tan⁻¹ |
| ∫ dx/√(x² − a²) | log |x + √(x² − a²)| + C | — |
| ∫ dx/√(x² + a²) | log |x + √(x² + a²)| + C | — |
| ∫ dx/√(a² − x²) | sin⁻¹(x/a) + C | root of a² − x² on the bottom → sin⁻¹ |
| ∫ √(a² − x²) dx | (x/2)√(a² − x²) + (a²/2) sin⁻¹(x/a) + C | — |
| ∫ √(x² + a²) dx | (x/2)√(x² + a²) + (a²/2) log |x + √(x² + a²)| + C | — |
| ∫ √(x² − a²) dx | (x/2)√(x² − a²) − (a²/2) log |x + √(x² − a²)| + C | note the minus in the middle |
Key Rule — the sign detector
Look at the sign inside the bracket, it tells you the answer type. A plus with no root (x² + a² on the bottom) gives tan⁻¹. A minus with the x² second under a root (√(a² − x²)) gives sin⁻¹. Anything else with a root gives a log. Two minutes memorising this saves you from guessing in the exam.
Example 3 — Straight from the table
Find ∫ dx/(x² + 9).
Compare with ∫ dx/(x² + a²). Here a² = 9, so a = 3.
∫ dx/(x² + 9) = (1/3) tan⁻¹(x/3) + C.
Check: d/dx[(1/3)tan⁻¹(x/3)] = (1/3) × (1/3)/(1 + x²/9) = (1/9) × 9/(9 + x²) = 1/(x² + 9). ✓
Compare with ∫ dx/(x² + a²). Here a² = 9, so a = 3.
∫ dx/(x² + 9) = (1/3) tan⁻¹(x/3) + C.
Check: d/dx[(1/3)tan⁻¹(x/3)] = (1/3) × (1/3)/(1 + x²/9) = (1/9) × 9/(9 + x²) = 1/(x² + 9). ✓
Common Mistake
Students write ∫ dx/x = log x + C and drop the modulus. It must be log |x|, because log x is undefined for negative x while 1/x is perfectly happy there. Similarly, never write ∫ xⁿ dx = xⁿ⁺¹/(n+1) when n = −1 — you would be dividing by zero. That single case is exactly why the log row exists.
Basic Rules And Rewriting Before You Integrate
Before any clever technique, there are two rules that let you break an integral into manageable pieces:
- Constants slide out: ∫ k·f(x) dx = k ∫ f(x) dx.
- Sums split up: ∫ [f(x) ± g(x)] dx = ∫ f(x) dx ± ∫ g(x) dx.
And then there is the rule that is not written in any textbook but wins the most marks: rewrite first, integrate second. A huge number of “hard” integrals are just algebra in disguise. Divide out the fraction, split the terms, apply a trigonometric identity — and suddenly every piece is in the standard table.
Common Mistake — there is no product rule for integration
∫ f(x)·g(x) dx is not equal to (∫ f dx)(∫ g dx). Likewise ∫ [f(x)/g(x)] dx is not (∫ f dx)/(∫ g dx). Integration splits happily over addition and pulls out constants — that is all. Products and quotients need substitution, by parts, or partial fractions.
Example 4 — Split it up, term by term
Find ∫ (x³⁄² + 5/x − 3 sec² x) dx.
Three separate table entries. Take them one at a time.
∫ x³⁄² dx = x⁵⁄²/(5/2) = (2/5) x⁵⁄²
∫ (5/x) dx = 5 log |x|
∫ 3 sec² x dx = 3 tan x
Answer: (2/5) x⁵⁄² + 5 log |x| − 3 tan x + C.
Three separate table entries. Take them one at a time.
∫ x³⁄² dx = x⁵⁄²/(5/2) = (2/5) x⁵⁄²
∫ (5/x) dx = 5 log |x|
∫ 3 sec² x dx = 3 tan x
Answer: (2/5) x⁵⁄² + 5 log |x| − 3 tan x + C.
Example 5 — Rewrite before you panic
Find ∫ (x² + 3x + 1)/x dx.
There is no formula for a quotient. But look — the bottom is a single term, so just divide each term on top by x:
(x² + 3x + 1)/x = x + 3 + 1/x
Now every piece is standard:
∫ (x + 3 + 1/x) dx = x²/2 + 3x + log |x| + C.
Answer: x²/2 + 3x + log |x| + C. The whole difficulty was algebra, not calculus.
There is no formula for a quotient. But look — the bottom is a single term, so just divide each term on top by x:
(x² + 3x + 1)/x = x + 3 + 1/x
Now every piece is standard:
∫ (x + 3 + 1/x) dx = x²/2 + 3x + log |x| + C.
Answer: x²/2 + 3x + log |x| + C. The whole difficulty was algebra, not calculus.
Example 6 — A trigonometric rewrite
Find ∫ (1 − sin x)/cos² x dx.
Split the fraction over the common denominator:
(1 − sin x)/cos² x = 1/cos² x − sin x/cos² x = sec² x − sec x tan x
Both are now table entries:
∫ sec² x dx = tan x and ∫ sec x tan x dx = sec x
Answer: tan x − sec x + C.
Check: d/dx(tan x − sec x) = sec² x − sec x tan x. ✓
Split the fraction over the common denominator:
(1 − sin x)/cos² x = 1/cos² x − sin x/cos² x = sec² x − sec x tan x
Both are now table entries:
∫ sec² x dx = tan x and ∫ sec x tan x dx = sec x
Answer: tan x − sec x + C.
Check: d/dx(tan x − sec x) = sec² x − sec x tan x. ✓
Exam Tip
When an integral looks impossible, spend thirty seconds asking “can I rewrite this?” before reaching for a technique. Divide out, split the fraction, rationalise, use sin² + cos² = 1, or multiply top and bottom by something useful. Rewriting is free marks; a wrong technique costs you five minutes.
Integration By Substitution
Substitution is the reverse of the chain rule, and it is the workhorse of this chapter. The idea is simple: if the integral is written in an awkward variable, change the variable to something friendlier.
Here is the everyday analogy. Suppose a recipe is written in cups and your measuring jug only has millilitres. You do not throw the recipe away — you convert the units, cook, and you are done. Substitution converts the “units” of the integral from x to a new letter t, in which the integral is something you already know.
Key Rule — the substitution method
To evaluate ∫ f(x) dx, put x = g(t), so that dx = g′(t) dt. Then ∫ f(x) dx = ∫ f(g(t)) g′(t) dt.
In practice you usually go the other way: you spot an inner function, set t = (that inner function), differentiate to get dt = (its derivative) dx, and swap. The substitution works only if the leftover part of the integrand is exactly that derivative, up to a constant factor. At the end, always substitute back so your answer is in terms of x.
In practice you usually go the other way: you spot an inner function, set t = (that inner function), differentiate to get dt = (its derivative) dx, and swap. The substitution works only if the leftover part of the integrand is exactly that derivative, up to a constant factor. At the end, always substitute back so your answer is in terms of x.
So how do you spot a substitution? Look for a function and its derivative living in the same integral. If you can see a chunk whose derivative is also sitting there (perhaps with a constant attached), that chunk is your t.
Example 7 — The classic pattern
Find ∫ 2x/(1 + x²) dx.
Spot it: the bottom is 1 + x², and its derivative is 2x — which is exactly what is on top.
Put t = 1 + x², so dt = 2x dx.
∫ 2x/(1 + x²) dx = ∫ dt/t = log |t| + C
Substitute back:
Answer: log(1 + x²) + C. (No modulus needed here — 1 + x² is always positive.)
Spot it: the bottom is 1 + x², and its derivative is 2x — which is exactly what is on top.
Put t = 1 + x², so dt = 2x dx.
∫ 2x/(1 + x²) dx = ∫ dt/t = log |t| + C
Substitute back:
Answer: log(1 + x²) + C. (No modulus needed here — 1 + x² is always positive.)
Example 8 — When only a constant is missing
Find ∫ x eˣ² dx. (Here the exponent is x².)
Put t = x². Then dt = 2x dx, so x dx = dt/2. The integral has x dx sitting in it — perfect.
∫ x eˣ² dx = ∫ eᵗ (dt/2) = (1/2) eᵗ + C
Answer: (1/2) eˣ² + C.
Check: d/dx[(1/2)eˣ²] = (1/2)·eˣ²·2x = x eˣ². ✓ The constant 1/2 is not optional — forgetting it is the most common slip in this whole section.
Put t = x². Then dt = 2x dx, so x dx = dt/2. The integral has x dx sitting in it — perfect.
∫ x eˣ² dx = ∫ eᵗ (dt/2) = (1/2) eᵗ + C
Answer: (1/2) eˣ² + C.
Check: d/dx[(1/2)eˣ²] = (1/2)·eˣ²·2x = x eˣ². ✓ The constant 1/2 is not optional — forgetting it is the most common slip in this whole section.
Example 9 — Proving a table entry
Show that ∫ tan x dx = log |sec x| + C.
Write tan x = sin x / cos x. Put t = cos x, so dt = − sin x dx, i.e. sin x dx = − dt.
∫ (sin x / cos x) dx = ∫ (− dt)/t = − log |t| + C = − log |cos x| + C
And − log |cos x| = log |cos x|⁻¹ = log |sec x|.
Answer: log |sec x| + C. Both forms are correct; write whichever the question hints at.
Write tan x = sin x / cos x. Put t = cos x, so dt = − sin x dx, i.e. sin x dx = − dt.
∫ (sin x / cos x) dx = ∫ (− dt)/t = − log |t| + C = − log |cos x| + C
And − log |cos x| = log |cos x|⁻¹ = log |sec x|.
Answer: log |sec x| + C. Both forms are correct; write whichever the question hints at.
Example 10 — One more, with a logarithm inside
Find ∫ (log x)/x dx.
The derivative of log x is 1/x, and 1/x is right there. So put t = log x, dt = dx/x.
∫ (log x)·(dx/x) = ∫ t dt = t²/2 + C
Answer: (log x)²/2 + C.
Check: d/dx[(log x)²/2] = (1/2)·2 log x · (1/x) = (log x)/x. ✓
The derivative of log x is 1/x, and 1/x is right there. So put t = log x, dt = dx/x.
∫ (log x)·(dx/x) = ∫ t dt = t²/2 + C
Answer: (log x)²/2 + C.
Check: d/dx[(log x)²/2] = (1/2)·2 log x · (1/x) = (log x)/x. ✓
Common Mistake
Never change the variable in the integrand but forget to change dx. Writing ∫ x eˣ² dx = ∫ eᵗ dx is meaningless — you now have two different variables in one integral. Every substitution has two halves: replace the function and replace dx. And after integrating in t, do not leave t in the final answer — substitute back to x.
Integrals Using Trigonometric Identities
You cannot integrate sin² x directly — it is not in the table. But you can integrate cos 2x. So the trick for trigonometric integrals is always the same: use an identity to convert powers and products into plain sines and cosines of multiple angles, which are all standard.
Key Rule — the four identities that do all the work
Powers → multiple angles:
sin² x = (1 − cos 2x)/2 cos² x = (1 + cos 2x)/2
sin³ x = (3 sin x − sin 3x)/4 cos³ x = (3 cos x + cos 3x)/4
Products → sums:
2 sin A cos B = sin(A + B) + sin(A − B)
2 cos A cos B = cos(A + B) + cos(A − B)
2 sin A sin B = cos(A − B) − cos(A + B)
Odd powers: if one of sin or cos appears to an odd power, peel off one factor and substitute using sin² + cos² = 1.
sin² x = (1 − cos 2x)/2 cos² x = (1 + cos 2x)/2
sin³ x = (3 sin x − sin 3x)/4 cos³ x = (3 cos x + cos 3x)/4
Products → sums:
2 sin A cos B = sin(A + B) + sin(A − B)
2 cos A cos B = cos(A + B) + cos(A − B)
2 sin A sin B = cos(A − B) − cos(A + B)
Odd powers: if one of sin or cos appears to an odd power, peel off one factor and substitute using sin² + cos² = 1.
Example 11 — The most-asked one in the chapter
Find ∫ sin² x dx.
Replace sin² x with (1 − cos 2x)/2:
∫ sin² x dx = (1/2) ∫ (1 − cos 2x) dx = (1/2)[x − (sin 2x)/2] + C
Answer: x/2 − (sin 2x)/4 + C.
Check: d/dx[x/2 − sin2x/4] = 1/2 − (2 cos 2x)/4 = 1/2 − (cos 2x)/2 = (1 − cos 2x)/2 = sin² x. ✓
Replace sin² x with (1 − cos 2x)/2:
∫ sin² x dx = (1/2) ∫ (1 − cos 2x) dx = (1/2)[x − (sin 2x)/2] + C
Answer: x/2 − (sin 2x)/4 + C.
Check: d/dx[x/2 − sin2x/4] = 1/2 − (2 cos 2x)/4 = 1/2 − (cos 2x)/2 = (1 − cos 2x)/2 = sin² x. ✓
Example 12 — A product of two sines/cosines
Find ∫ sin 3x cos x dx.
Use 2 sin A cos B = sin(A + B) + sin(A − B) with A = 3x, B = x:
sin 3x cos x = (1/2)[sin 4x + sin 2x]
∫ sin 3x cos x dx = (1/2)[−(cos 4x)/4 − (cos 2x)/2] + C
Answer: − (cos 4x)/8 − (cos 2x)/4 + C.
Use 2 sin A cos B = sin(A + B) + sin(A − B) with A = 3x, B = x:
sin 3x cos x = (1/2)[sin 4x + sin 2x]
∫ sin 3x cos x dx = (1/2)[−(cos 4x)/4 − (cos 2x)/2] + C
Answer: − (cos 4x)/8 − (cos 2x)/4 + C.
Example 13 — An odd power — peel and substitute
Find ∫ cos³ x dx.
Cosine has an odd power, so peel off one cos x and convert the rest:
cos³ x = cos² x · cos x = (1 − sin² x) cos x
Now put t = sin x, dt = cos x dx:
∫ (1 − t²) dt = t − t³/3 + C
Answer: sin x − (sin³ x)/3 + C.
Check: d/dx = cos x − 3sin²x cos x/3 = cos x(1 − sin² x) = cos³ x. ✓
Cosine has an odd power, so peel off one cos x and convert the rest:
cos³ x = cos² x · cos x = (1 − sin² x) cos x
Now put t = sin x, dt = cos x dx:
∫ (1 − t²) dt = t − t³/3 + C
Answer: sin x − (sin³ x)/3 + C.
Check: d/dx = cos x − 3sin²x cos x/3 = cos x(1 − sin² x) = cos³ x. ✓
Example 14 — A product of two even powers
Find ∫ sin² x cos² x dx.
Do not expand blindly — use sin x cos x = (sin 2x)/2 first:
sin² x cos² x = (sin x cos x)² = (sin² 2x)/4
Then apply the power identity again to sin² 2x = (1 − cos 4x)/2:
= (1/4)·(1 − cos 4x)/2 = (1 − cos 4x)/8
∫ (1 − cos 4x)/8 dx = x/8 − (sin 4x)/32 + C
Answer: x/8 − (sin 4x)/32 + C.
Do not expand blindly — use sin x cos x = (sin 2x)/2 first:
sin² x cos² x = (sin x cos x)² = (sin² 2x)/4
Then apply the power identity again to sin² 2x = (1 − cos 4x)/2:
= (1/4)·(1 − cos 4x)/2 = (1 − cos 4x)/8
∫ (1 − cos 4x)/8 dx = x/8 − (sin 4x)/32 + C
Answer: x/8 − (sin 4x)/32 + C.
Exam Tip
Nearly every trigonometric integral in this chapter is solved by one of three moves: (1) an identity that lowers a power, (2) a product-to-sum formula, or (3) peeling off one factor from an odd power and substituting. Before you write anything, ask which of the three it is. That single question replaces a lot of trial and error.
Integration By Partial Fractions
Some fractions are impossible to integrate as one lump but trivial once you break them apart. Think of it like paying a bill of ₹175 — nobody has a ₹175 note, but a ₹100 plus a ₹50 plus a ₹20 plus a ₹5 does the job. Partial fractions breaks one hostile fraction into a sum of small friendly ones.
Before anything else, check the degrees. Partial fractions only works on a proper fraction, where the degree of the numerator is strictly less than the degree of the denominator. If it is improper, divide first and then decompose the remainder.
Your syllabus lists four standard forms. Here they are, with the shape you must assume:
| Form of the fraction | What it is called | Assume this form |
|---|---|---|
| (px + q)/((x − a)(x − b)), a ≠ b | Distinct linear factors | A/(x − a) + B/(x − b) |
| (px + q)/(x − a)² | Repeated linear factor | A/(x − a) + B/(x − a)² |
| (px² + qx + r)/((x − a)(x − b)(x − c)) | Three distinct linear factors | A/(x − a) + B/(x − b) + C/(x − c) |
| (px² + qx + r)/((x − a)(x² + bx + c)) | An irreducible quadratic factor | A/(x − a) + (Bx + C)/(x² + bx + c) |
Key Rule — the shape of the numerator
Over a linear factor you put a constant (A). Over an irreducible quadratic factor you put a linear expression (Bx + C). And a factor repeated k times needs one term for every power from 1 up to k. Getting this shape right is half the question; the rest is solving for the letters.
Example 15 — Case 1 — distinct linear factors
Find ∫ dx/((x − 1)(x + 2)).
Assume 1/((x − 1)(x + 2)) = A/(x − 1) + B/(x + 2).
Multiply through by (x − 1)(x + 2): 1 = A(x + 2) + B(x − 1).
Now choose clever values of x to kill one letter at a time:
Put x = 1: 1 = A(3) → A = 1/3
Put x = −2: 1 = B(−3) → B = −1/3
∫ [ (1/3)/(x − 1) − (1/3)/(x + 2) ] dx = (1/3) log |x − 1| − (1/3) log |x + 2| + C
Answer: (1/3) log |(x − 1)/(x + 2)| + C.
Assume 1/((x − 1)(x + 2)) = A/(x − 1) + B/(x + 2).
Multiply through by (x − 1)(x + 2): 1 = A(x + 2) + B(x − 1).
Now choose clever values of x to kill one letter at a time:
Put x = 1: 1 = A(3) → A = 1/3
Put x = −2: 1 = B(−3) → B = −1/3
∫ [ (1/3)/(x − 1) − (1/3)/(x + 2) ] dx = (1/3) log |x − 1| − (1/3) log |x + 2| + C
Answer: (1/3) log |(x − 1)/(x + 2)| + C.
Example 16 — Case 2 — a repeated linear factor
Find ∫ (3x + 2)/((x − 1)(x + 1)²) dx.
The factor (x + 1) is repeated, so we need two terms for it:
(3x + 2)/((x − 1)(x + 1)²) = A/(x − 1) + B/(x + 1) + C₁/(x + 1)²
Multiply up: 3x + 2 = A(x + 1)² + B(x − 1)(x + 1) + C₁(x − 1)
Put x = 1: 5 = A(4) → A = 5/4
Put x = −1: −1 = C₁(−2) → C₁ = 1/2
Compare x² coefficients: 0 = A + B → B = −5/4
So the integral is ∫ [ (5/4)/(x − 1) − (5/4)/(x + 1) + (1/2)/(x + 1)² ] dx.
The third piece: ∫ (1/2)(x + 1)⁻² dx = (1/2)·(x + 1)⁻¹/(−1) = − 1/(2(x + 1)).
Answer: (5/4) log |x − 1| − (5/4) log |x + 1| − 1/(2(x + 1)) + C.
Watch that minus sign on the last term — integrating (x + 1)⁻² flips the sign. It is the single most common error in this example.
The factor (x + 1) is repeated, so we need two terms for it:
(3x + 2)/((x − 1)(x + 1)²) = A/(x − 1) + B/(x + 1) + C₁/(x + 1)²
Multiply up: 3x + 2 = A(x + 1)² + B(x − 1)(x + 1) + C₁(x − 1)
Put x = 1: 5 = A(4) → A = 5/4
Put x = −1: −1 = C₁(−2) → C₁ = 1/2
Compare x² coefficients: 0 = A + B → B = −5/4
So the integral is ∫ [ (5/4)/(x − 1) − (5/4)/(x + 1) + (1/2)/(x + 1)² ] dx.
The third piece: ∫ (1/2)(x + 1)⁻² dx = (1/2)·(x + 1)⁻¹/(−1) = − 1/(2(x + 1)).
Answer: (5/4) log |x − 1| − (5/4) log |x + 1| − 1/(2(x + 1)) + C.
Watch that minus sign on the last term — integrating (x + 1)⁻² flips the sign. It is the single most common error in this example.
Example 17 — Case 3 — an irreducible quadratic factor
Find ∫ (x² + 1)/((x + 1)(x² + 4)) dx.
x² + 4 never factorises over the reals, so it gets a linear numerator:
(x² + 1)/((x + 1)(x² + 4)) = A/(x + 1) + (Bx + C₁)/(x² + 4)
Multiply up: x² + 1 = A(x² + 4) + (Bx + C₁)(x + 1)
Put x = −1: 2 = A(5) → A = 2/5
Compare x²: 1 = A + B → B = 3/5
Compare constants: 1 = 4A + C₁ → 1 = 8/5 + C₁ → C₁ = −3/5
So we need ∫ (2/5)/(x + 1) dx + ∫ (3/5)(x − 1)/(x² + 4) dx.
Split the second piece into (3/5)·x/(x²+4) and −(3/5)·1/(x²+4):
∫ x/(x² + 4) dx = (1/2) log(x² + 4) and ∫ dx/(x² + 4) = (1/2) tan⁻¹(x/2)
Answer: (2/5) log |x + 1| + (3/10) log(x² + 4) − (3/10) tan⁻¹(x/2) + C.
x² + 4 never factorises over the reals, so it gets a linear numerator:
(x² + 1)/((x + 1)(x² + 4)) = A/(x + 1) + (Bx + C₁)/(x² + 4)
Multiply up: x² + 1 = A(x² + 4) + (Bx + C₁)(x + 1)
Put x = −1: 2 = A(5) → A = 2/5
Compare x²: 1 = A + B → B = 3/5
Compare constants: 1 = 4A + C₁ → 1 = 8/5 + C₁ → C₁ = −3/5
So we need ∫ (2/5)/(x + 1) dx + ∫ (3/5)(x − 1)/(x² + 4) dx.
Split the second piece into (3/5)·x/(x²+4) and −(3/5)·1/(x²+4):
∫ x/(x² + 4) dx = (1/2) log(x² + 4) and ∫ dx/(x² + 4) = (1/2) tan⁻¹(x/2)
Answer: (2/5) log |x + 1| + (3/10) log(x² + 4) − (3/10) tan⁻¹(x/2) + C.
Example 18 — A shortcut worth knowing
Find ∫ x/((x² + 1)(x² + 3)) dx.
Do not decompose in x — notice that x dx is present and everything else involves x². Substitute first: put t = x², so dt = 2x dx, i.e. x dx = dt/2.
The integral becomes (1/2) ∫ dt/((t + 1)(t + 3)) — now a Case 1 partial fraction in t.
1/((t + 1)(t + 3)) = (1/2)/(t + 1) − (1/2)/(t + 3)
So we get (1/2)·(1/2)[log|t + 1| − log|t + 3|] = (1/4) log((t+1)/(t+3)).
Answer: (1/4) log[(x² + 1)/(x² + 3)] + C.
Substitution and partial fractions are friends, not rivals. Whenever every power of x is even and you have a lone x dx, try t = x² first.
Do not decompose in x — notice that x dx is present and everything else involves x². Substitute first: put t = x², so dt = 2x dx, i.e. x dx = dt/2.
The integral becomes (1/2) ∫ dt/((t + 1)(t + 3)) — now a Case 1 partial fraction in t.
1/((t + 1)(t + 3)) = (1/2)/(t + 1) − (1/2)/(t + 3)
So we get (1/2)·(1/2)[log|t + 1| − log|t + 3|] = (1/4) log((t+1)/(t+3)).
Answer: (1/4) log[(x² + 1)/(x² + 3)] + C.
Substitution and partial fractions are friends, not rivals. Whenever every power of x is even and you have a lone x dx, try t = x² first.
Common Mistake
Two traps. First, decomposing an improper fraction — if the top degree is greater than or equal to the bottom degree, do the division first. Second, putting a plain constant over a quadratic factor. Writing A/(x² + 4) instead of (Bx + C)/(x² + 4) throws away a whole degree of freedom and your equations will refuse to solve.
Integration By Parts
LIATE: a reliable order of preference for picking the first function in integration by parts.
Substitution handles a function inside another function. But what about a genuine product of two unrelated functions, like x eˣ or x² log x? Substitution gets you nowhere, because neither factor is the derivative of the other. That is where integration by parts comes in — it is the reverse of the product rule.
Key Rule — the by-parts formula
∫ u · v dx = u ∫ v dx − ∫ [ (du/dx) · ∫ v dx ] dx
In words: (first function) × (integral of second) − integral of [ (derivative of first) × (integral of second) ].
Choosing which factor is “first” (u) decides whether the method helps or makes things worse. Use LIATE: whichever type comes earlier in Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential becomes u. The whole point is that differentiating u should make it simpler.
In words: (first function) × (integral of second) − integral of [ (derivative of first) × (integral of second) ].
Choosing which factor is “first” (u) decides whether the method helps or makes things worse. Use LIATE: whichever type comes earlier in Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential becomes u. The whole point is that differentiating u should make it simpler.
Example 19 — The starter example
Find ∫ x eˣ dx.
By LIATE, x is Algebraic (A) and eˣ is Exponential (E). A comes first, so u = x and the second function is eˣ.
∫ x eˣ dx = x · eˣ − ∫ [ 1 · eˣ ] dx = x eˣ − eˣ + C
Answer: (x − 1) eˣ + C.
Check: d/dx[(x − 1)eˣ] = eˣ + (x − 1)eˣ = x eˣ. ✓
Notice why it worked: differentiating x gave 1, which killed the algebraic part entirely. Had we chosen u = eˣ instead, we would have got ∫ (x²/2) eˣ dx — strictly worse.
By LIATE, x is Algebraic (A) and eˣ is Exponential (E). A comes first, so u = x and the second function is eˣ.
∫ x eˣ dx = x · eˣ − ∫ [ 1 · eˣ ] dx = x eˣ − eˣ + C
Answer: (x − 1) eˣ + C.
Check: d/dx[(x − 1)eˣ] = eˣ + (x − 1)eˣ = x eˣ. ✓
Notice why it worked: differentiating x gave 1, which killed the algebraic part entirely. Had we chosen u = eˣ instead, we would have got ∫ (x²/2) eˣ dx — strictly worse.
Example 20 — When a logarithm is present
Find ∫ x² log x dx.
L beats A, so u = log x and the second function is x².
∫ x² log x dx = log x · (x³/3) − ∫ [ (1/x) · (x³/3) ] dx
= (x³ log x)/3 − (1/3) ∫ x² dx
= (x³ log x)/3 − x³/9 + C
Answer: (x³ log x)/3 − x³/9 + C.
Check: d/dx = x² log x + (x³/3)(1/x) − 3x²/9 = x² log x + x²/3 − x²/3 = x² log x. ✓
L beats A, so u = log x and the second function is x².
∫ x² log x dx = log x · (x³/3) − ∫ [ (1/x) · (x³/3) ] dx
= (x³ log x)/3 − (1/3) ∫ x² dx
= (x³ log x)/3 − x³/9 + C
Answer: (x³ log x)/3 − x³/9 + C.
Check: d/dx = x² log x + (x³/3)(1/x) − 3x²/9 = x² log x + x²/3 − x²/3 = x² log x. ✓
Example 21 — The clever one — a “single” function
Find ∫ log x dx.
This looks like it has only one function, so by parts seems impossible. The trick: write it as log x × 1. Now u = log x (L) and the second function is 1.
∫ log x · 1 dx = log x · x − ∫ [ (1/x) · x ] dx = x log x − ∫ 1 dx
Answer: x log x − x + C.
The same “multiply by 1” trick gives you ∫ sin⁻¹x dx, ∫ tan⁻¹x dx and every other inverse-trigonometric integral. Keep it in your pocket.
This looks like it has only one function, so by parts seems impossible. The trick: write it as log x × 1. Now u = log x (L) and the second function is 1.
∫ log x · 1 dx = log x · x − ∫ [ (1/x) · x ] dx = x log x − ∫ 1 dx
Answer: x log x − x + C.
The same “multiply by 1” trick gives you ∫ sin⁻¹x dx, ∫ tan⁻¹x dx and every other inverse-trigonometric integral. Keep it in your pocket.
Example 22 — Applying by parts twice
Find ∫ x² eˣ dx.
u = x² (A beats E), second function eˣ:
∫ x² eˣ dx = x² eˣ − ∫ 2x eˣ dx
The leftover integral is Example 19 × 2, which we already know equals 2(x − 1)eˣ.
= x² eˣ − 2(x − 1)eˣ = (x² − 2x + 2) eˣ + C
Answer: (x² − 2x + 2) eˣ + C.
Rule of thumb: an xⁿ multiplying eˣ or sin/cos needs by parts applied n times — each round drops the power by one.
u = x² (A beats E), second function eˣ:
∫ x² eˣ dx = x² eˣ − ∫ 2x eˣ dx
The leftover integral is Example 19 × 2, which we already know equals 2(x − 1)eˣ.
= x² eˣ − 2(x − 1)eˣ = (x² − 2x + 2) eˣ + C
Answer: (x² − 2x + 2) eˣ + C.
Rule of thumb: an xⁿ multiplying eˣ or sin/cos needs by parts applied n times — each round drops the power by one.
Example 23 — The “boomerang” integral
Find I = ∫ eˣ sin x dx.
Neither factor simplifies on differentiating, so by parts seems to go in circles. It does — and that is exactly the trick. Take u = sin x, second function eˣ:
I = eˣ sin x − ∫ eˣ cos x dx
Apply by parts again to the new integral (u = cos x):
∫ eˣ cos x dx = eˣ cos x + ∫ eˣ sin x dx = eˣ cos x + I
Substitute back:
I = eˣ sin x − eˣ cos x − I → 2I = eˣ(sin x − cos x)
Answer: I = eˣ(sin x − cos x)/2 + C.
When the original integral reappears, do not despair — move it to the left side and divide. Just be sure to use the same choice of u both times, or everything cancels to 0 = 0.
Neither factor simplifies on differentiating, so by parts seems to go in circles. It does — and that is exactly the trick. Take u = sin x, second function eˣ:
I = eˣ sin x − ∫ eˣ cos x dx
Apply by parts again to the new integral (u = cos x):
∫ eˣ cos x dx = eˣ cos x + ∫ eˣ sin x dx = eˣ cos x + I
Substitute back:
I = eˣ sin x − eˣ cos x − I → 2I = eˣ(sin x − cos x)
Answer: I = eˣ(sin x − cos x)/2 + C.
When the original integral reappears, do not despair — move it to the left side and divide. Just be sure to use the same choice of u both times, or everything cancels to 0 = 0.
Common Mistake
After the first by-parts step, students often forget that the whole second term is subtracted, including its own sign. If the remaining integral evaluates to a negative quantity, subtracting it makes the term positive. Write brackets around the second integral before you evaluate it — − ∫ [ … ] dx — and the sign will look after itself.
The Special Result For eˣ[f(x) + f′(x)]
This is a small result with a big payoff. It is named explicitly in your syllabus, and questions built on it appear regularly — usually disguised, so that the whole difficulty is recognising it.
Key Rule
∫ eˣ [ f(x) + f′(x) ] dx = eˣ f(x) + C
Why it is true: by the product rule, d/dx[eˣ f(x)] = eˣ f(x) + eˣ f′(x) = eˣ[f(x) + f′(x)]. Integrating both sides gives the result immediately.
More generally, ∫ eax[a f(x) + f′(x)] dx = eax f(x) + C.
Why it is true: by the product rule, d/dx[eˣ f(x)] = eˣ f(x) + eˣ f′(x) = eˣ[f(x) + f′(x)]. Integrating both sides gives the result immediately.
More generally, ∫ eax[a f(x) + f′(x)] dx = eax f(x) + C.
The whole skill is pattern-spotting. When you see eˣ multiplying a bracket containing two terms, ask: is one of these terms the derivative of the other? If yes, the answer is eˣ times the term that is not the derivative.
Example 24 — Spotting the pattern
Find ∫ eˣ (log x + 1/x) dx.
Test the bracket. Is 1/x the derivative of log x? Yes — exactly.
So f(x) = log x and f′(x) = 1/x. The pattern fits perfectly.
Answer: eˣ log x + C.
Check: d/dx[eˣ log x] = eˣ log x + eˣ/x = eˣ(log x + 1/x). ✓ No by-parts, no working — just recognition.
Test the bracket. Is 1/x the derivative of log x? Yes — exactly.
So f(x) = log x and f′(x) = 1/x. The pattern fits perfectly.
Answer: eˣ log x + C.
Check: d/dx[eˣ log x] = eˣ log x + eˣ/x = eˣ(log x + 1/x). ✓ No by-parts, no working — just recognition.
Example 25 — When you must create the pattern
Find ∫ eˣ (x + 1)/(x + 2)² dx.
The bracket is a single fraction, so first split it. Write the numerator to match the denominator:
(x + 1) = (x + 2) − 1
So (x + 1)/(x + 2)² = (x + 2)/(x + 2)² − 1/(x + 2)² = 1/(x + 2) − 1/(x + 2)²
Now take f(x) = 1/(x + 2). Then f′(x) = −1/(x + 2)², which is precisely the second term.
The integrand is eˣ[f(x) + f′(x)].
Answer: eˣ/(x + 2) + C.
Check: d/dx[eˣ(x+2)⁻¹] = eˣ(x+2)⁻¹ − eˣ(x+2)⁻² = eˣ[(x+2) − 1]/(x+2)² = eˣ(x+1)/(x+2)². ✓
The bracket is a single fraction, so first split it. Write the numerator to match the denominator:
(x + 1) = (x + 2) − 1
So (x + 1)/(x + 2)² = (x + 2)/(x + 2)² − 1/(x + 2)² = 1/(x + 2) − 1/(x + 2)²
Now take f(x) = 1/(x + 2). Then f′(x) = −1/(x + 2)², which is precisely the second term.
The integrand is eˣ[f(x) + f′(x)].
Answer: eˣ/(x + 2) + C.
Check: d/dx[eˣ(x+2)⁻¹] = eˣ(x+2)⁻¹ − eˣ(x+2)⁻² = eˣ[(x+2) − 1]/(x+2)² = eˣ(x+1)/(x+2)². ✓
Example 26 — A trigonometric version
Find ∫ eˣ (1/x − 1/x²) dx.
Take f(x) = 1/x. Then f′(x) = −1/x². The bracket is f(x) + f′(x) exactly.
Answer: eˣ/x + C.
Check: d/dx[eˣ/x] = eˣ/x − eˣ/x² = eˣ(1/x − 1/x²). ✓
Note how a minus sign inside the bracket is not a problem — it is simply part of f′(x). Do not reject the pattern just because you see a subtraction.
Take f(x) = 1/x. Then f′(x) = −1/x². The bracket is f(x) + f′(x) exactly.
Answer: eˣ/x + C.
Check: d/dx[eˣ/x] = eˣ/x − eˣ/x² = eˣ(1/x − 1/x²). ✓
Note how a minus sign inside the bracket is not a problem — it is simply part of f′(x). Do not reject the pattern just because you see a subtraction.
Exam Tip
If the question gives you eˣ times a bracket and the pattern does not fit immediately, try three moves before giving up: split the fraction (Example 25), pull out a constant, or rewrite the trigonometric part with a half-angle identity. Examiners deliberately hide this pattern one algebraic step away from the surface.
Special Type Integrals: The a² And x² Family
Your syllabus names a specific list of integral types and says “problems based on them”. This section and the three that follow work through that list in order. Start here, with the clean cases where the quadratic is already a difference or sum of two squares.
These four rows of the standard table are the core:
- ∫ dx/(x² − a²) = (1/2a) log |(x − a)/(x + a)| + C
- ∫ dx/(a² − x²) = (1/2a) log |(a + x)/(a − x)| + C
- ∫ dx/(x² + a²) = (1/a) tan⁻¹(x/a) + C
- ∫ dx/√(x² ± a²) = log |x + √(x² ± a²)| + C and ∫ dx/√(a² − x²) = sin⁻¹(x/a) + C
Key Rule — identify a, then substitute into the formula
Every question of this type is three steps. (1) Write the constant as a perfect square to read off a. (2) Match the sign pattern to the correct row. (3) Substitute. If the coefficient of x² is not 1, factor it out of the whole denominator first — that constant then comes outside the integral.
Example 27 — Reading off a
Find ∫ dx/(x² − 16).
16 = 4², so a = 4, and the pattern is x² − a².
∫ dx/(x² − 16) = (1/8) log |(x − 4)/(x + 4)| + C.
Sanity check by partial fractions: 1/((x−4)(x+4)) = (1/8)/(x−4) − (1/8)/(x+4), which integrates to (1/8)[log|x−4| − log|x+4|]. Same answer. ✓ The formula is just partial fractions done once and remembered.
16 = 4², so a = 4, and the pattern is x² − a².
∫ dx/(x² − 16) = (1/8) log |(x − 4)/(x + 4)| + C.
Sanity check by partial fractions: 1/((x−4)(x+4)) = (1/8)/(x−4) − (1/8)/(x+4), which integrates to (1/8)[log|x−4| − log|x+4|]. Same answer. ✓ The formula is just partial fractions done once and remembered.
Example 28 — A root on the bottom
Find ∫ dx/√(x² + 16).
Here a = 4 and the pattern is √(x² + a²).
Answer: log |x + √(x² + 16)| + C.
Compare with ∫ dx/√(9 − x²), where the x² is subtracted from the constant. That is the sin⁻¹ row, giving sin⁻¹(x/3) + C. The order of the two terms under the root is the whole difference between a log answer and a sin⁻¹ answer — read it carefully every single time.
Here a = 4 and the pattern is √(x² + a²).
Answer: log |x + √(x² + 16)| + C.
Compare with ∫ dx/√(9 − x²), where the x² is subtracted from the constant. That is the sin⁻¹ row, giving sin⁻¹(x/3) + C. The order of the two terms under the root is the whole difference between a log answer and a sin⁻¹ answer — read it carefully every single time.
Example 29 — When the coefficient of x² is not 1
Find ∫ dx/(4x² + 9).
Factor 4 out of the denominator:
4x² + 9 = 4(x² + 9/4)
∫ dx/(4x² + 9) = (1/4) ∫ dx/(x² + (3/2)²) = (1/4) · (1/(3/2)) tan⁻¹(x/(3/2)) + C
= (1/4)(2/3) tan⁻¹(2x/3) + C
Answer: (1/6) tan⁻¹(2x/3) + C.
Check: d/dx[(1/6)tan⁻¹(2x/3)] = (1/6)·(2/3)/(1 + 4x²/9) = (1/9)·9/(9 + 4x²) = 1/(4x² + 9). ✓
Factor 4 out of the denominator:
4x² + 9 = 4(x² + 9/4)
∫ dx/(4x² + 9) = (1/4) ∫ dx/(x² + (3/2)²) = (1/4) · (1/(3/2)) tan⁻¹(x/(3/2)) + C
= (1/4)(2/3) tan⁻¹(2x/3) + C
Answer: (1/6) tan⁻¹(2x/3) + C.
Check: d/dx[(1/6)tan⁻¹(2x/3)] = (1/6)·(2/3)/(1 + 4x²/9) = (1/9)·9/(9 + 4x²) = 1/(4x² + 9). ✓
Example 30 — A root with a coefficient
Find ∫ dx/√(4 − 9x²).
Factor 9 out from under the root:
√(4 − 9x²) = √(9(4/9 − x²)) = 3√((2/3)² − x²)
∫ dx/√(4 − 9x²) = (1/3) ∫ dx/√((2/3)² − x²) = (1/3) sin⁻¹(x/(2/3)) + C
Answer: (1/3) sin⁻¹(3x/2) + C.
Factor 9 out from under the root:
√(4 − 9x²) = √(9(4/9 − x²)) = 3√((2/3)² − x²)
∫ dx/√(4 − 9x²) = (1/3) ∫ dx/√((2/3)² − x²) = (1/3) sin⁻¹(x/(2/3)) + C
Answer: (1/3) sin⁻¹(3x/2) + C.
Common Mistake
∫ dx/(x² − a²) and ∫ dx/(a² − x²) are not the same up to a sign of the answer — the bracket contents flip too. For x² − a² the log is |(x − a)/(x + a)|; for a² − x² it is |(a + x)/(a − x)|. Memorise it as “whichever square is written first goes first inside the bracket.”
Completing The Square For ax² + bx + c
Now the general case: ∫ dx/(ax² + bx + c) and ∫ dx/√(ax² + bx + c). The middle term bx is the only thing standing between you and the previous section — and completing the square makes it disappear.
The idea is genuinely simple. A quadratic with a middle term is just a perfect square that has been shifted sideways. Completing the square finds the shift, and once the quadratic reads (something)² ± (constant)², you are back in the standard table with x replaced by that “something”.
Key Rule — the three-step routine
Step 1. If the coefficient of x² is not 1, take it out as a common factor from the whole quadratic.
Step 2. Complete the square: x² + bx = (x + b/2)² − (b/2)². Half the middle coefficient, square it, add and subtract it.
Step 3. Substitute t = x + b/2 (so dt = dx) and apply the matching standard formula. Then put x back.
Because dt = dx exactly, you can in practice skip writing the substitution and just treat (x + b/2) as the new “x”.
Step 2. Complete the square: x² + bx = (x + b/2)² − (b/2)². Half the middle coefficient, square it, add and subtract it.
Step 3. Substitute t = x + b/2 (so dt = dx) and apply the matching standard formula. Then put x back.
Because dt = dx exactly, you can in practice skip writing the substitution and just treat (x + b/2) as the new “x”.
Example 31 — A tan⁻¹ case
Find ∫ dx/(x² + 4x + 13).
Complete the square. Half of 4 is 2, and 2² = 4:
x² + 4x + 13 = (x² + 4x + 4) + 9 = (x + 2)² + 3²
Now it is the x² + a² row with a = 3 and x replaced by (x + 2):
Answer: (1/3) tan⁻¹((x + 2)/3) + C.
Check: d/dx = (1/3)·(1/3)/(1 + (x+2)²/9) = 1/(9 + (x+2)²) = 1/(x² + 4x + 13). ✓
Complete the square. Half of 4 is 2, and 2² = 4:
x² + 4x + 13 = (x² + 4x + 4) + 9 = (x + 2)² + 3²
Now it is the x² + a² row with a = 3 and x replaced by (x + 2):
Answer: (1/3) tan⁻¹((x + 2)/3) + C.
Check: d/dx = (1/3)·(1/3)/(1 + (x+2)²/9) = 1/(9 + (x+2)²) = 1/(x² + 4x + 13). ✓
Example 32 — A log case under a root
Find ∫ dx/√(x² + 6x + 5).
Half of 6 is 3, and 3² = 9:
x² + 6x + 5 = (x² + 6x + 9) − 4 = (x + 3)² − 2²
This is the √(x² − a²) row with a = 2:
Answer: log |(x + 3) + √((x + 3)² − 4)| + C, which you may also leave as log |(x + 3) + √(x² + 6x + 5)| + C.
Half of 6 is 3, and 3² = 9:
x² + 6x + 5 = (x² + 6x + 9) − 4 = (x + 3)² − 2²
This is the √(x² − a²) row with a = 2:
Answer: log |(x + 3) + √((x + 3)² − 4)| + C, which you may also leave as log |(x + 3) + √(x² + 6x + 5)| + C.
Example 33 — When the x² term is negative
Find ∫ dx/√(5 − 4x − x²).
First arrange it so the x² is grouped with the x term, with a minus outside:
5 − 4x − x² = 5 − (x² + 4x) = 5 − [(x + 2)² − 4] = 9 − (x + 2)²
So it is 3² − (x + 2)², the sin⁻¹ row with a = 3:
Answer: sin⁻¹((x + 2)/3) + C.
Check: d/dx = (1/3)/√(1 − (x+2)²/9) = 1/√(9 − (x+2)²) = 1/√(5 − 4x − x²). ✓
First arrange it so the x² is grouped with the x term, with a minus outside:
5 − 4x − x² = 5 − (x² + 4x) = 5 − [(x + 2)² − 4] = 9 − (x + 2)²
So it is 3² − (x + 2)², the sin⁻¹ row with a = 3:
Answer: sin⁻¹((x + 2)/3) + C.
Check: d/dx = (1/3)/√(1 − (x+2)²/9) = 1/√(9 − (x+2)²) = 1/√(5 − 4x − x²). ✓
Example 34 — With a coefficient in front
Find ∫ dx/(2x² − 4x + 10).
Take 2 out first:
2x² − 4x + 10 = 2(x² − 2x + 5) = 2[(x − 1)² + 4]
∫ dx/(2x² − 4x + 10) = (1/2) ∫ dx/((x − 1)² + 2²) = (1/2)·(1/2) tan⁻¹((x − 1)/2) + C
Answer: (1/4) tan⁻¹((x − 1)/2) + C.
Take 2 out first:
2x² − 4x + 10 = 2(x² − 2x + 5) = 2[(x − 1)² + 4]
∫ dx/(2x² − 4x + 10) = (1/2) ∫ dx/((x − 1)² + 2²) = (1/2)·(1/2) tan⁻¹((x − 1)/2) + C
Answer: (1/4) tan⁻¹((x − 1)/2) + C.
Common Mistake
Two sign traps. When the coefficient of x² is negative, you must factor the minus out of the x² and the x term together — forgetting the x term is what produces wrong answers in Example 33. And when you add (b/2)² to complete the square, you must subtract it again. Adding without subtracting silently changes the question.
Integrals Of The Type (px + q) Over A Quadratic
Next on the syllabus list: ∫ (px + q)/(ax² + bx + c) dx and ∫ (px + q)/√(ax² + bx + c) dx. A linear numerator on top of a quadratic. There is one beautiful trick that handles both, and once you see it you will never forget it.
The trick is to split the numerator into two useful halves: a piece that is exactly a multiple of the derivative of the denominator (which integrates instantly by substitution), and a leftover constant (which becomes a completing-the-square problem you already solved in the last section).
Key Rule — the split
Write px + q = λ · d/dx(ax² + bx + c) + μ, that is, px + q = λ(2ax + b) + μ.
Compare coefficients: λ = p/(2a), and then μ = q − λb.
The integral now splits into two:
• λ ∫ (2ax + b)/(quadratic) dx — pure substitution, t = the quadratic. Gives a log (or a root).
• μ ∫ dx/(quadratic) — complete the square, use the standard table.
Add the two results. That is the entire method.
Compare coefficients: λ = p/(2a), and then μ = q − λb.
The integral now splits into two:
• λ ∫ (2ax + b)/(quadratic) dx — pure substitution, t = the quadratic. Gives a log (or a root).
• μ ∫ dx/(quadratic) — complete the square, use the standard table.
Add the two results. That is the entire method.
Example 35 — The standard version
Find ∫ (2x + 3)/(x² + 4x + 8) dx.
The derivative of the denominator is 2x + 4. Write the numerator in terms of it:
2x + 3 = 1·(2x + 4) − 1
So the integral splits as
∫ (2x + 4)/(x² + 4x + 8) dx − ∫ dx/(x² + 4x + 8)
First piece: put t = x² + 4x + 8, dt = (2x + 4) dx, giving ∫ dt/t = log |t| = log(x² + 4x + 8).
Second piece: x² + 4x + 8 = (x + 2)² + 2², so ∫ dx/((x+2)² + 4) = (1/2) tan⁻¹((x + 2)/2).
Answer: log(x² + 4x + 8) − (1/2) tan⁻¹((x + 2)/2) + C.
Check: differentiating gives (2x+4)/(x²+4x+8) − (1/2)·(1/2)/(1 + (x+2)²/4) = (2x+4)/(x²+4x+8) − 1/(x²+4x+8) = (2x+3)/(x²+4x+8). ✓
The derivative of the denominator is 2x + 4. Write the numerator in terms of it:
2x + 3 = 1·(2x + 4) − 1
So the integral splits as
∫ (2x + 4)/(x² + 4x + 8) dx − ∫ dx/(x² + 4x + 8)
First piece: put t = x² + 4x + 8, dt = (2x + 4) dx, giving ∫ dt/t = log |t| = log(x² + 4x + 8).
Second piece: x² + 4x + 8 = (x + 2)² + 2², so ∫ dx/((x+2)² + 4) = (1/2) tan⁻¹((x + 2)/2).
Answer: log(x² + 4x + 8) − (1/2) tan⁻¹((x + 2)/2) + C.
Check: differentiating gives (2x+4)/(x²+4x+8) − (1/2)·(1/2)/(1 + (x+2)²/4) = (2x+4)/(x²+4x+8) − 1/(x²+4x+8) = (2x+3)/(x²+4x+8). ✓
Example 36 — With a fraction for λ
Find ∫ (3x − 2)/(x² − 2x + 5) dx.
Derivative of the denominator is 2x − 2. Set 3x − 2 = λ(2x − 2) + μ.
Comparing x terms: 3 = 2λ → λ = 3/2.
Comparing constants: −2 = −2λ + μ = −3 + μ → μ = 1.
∫ (3x − 2)/(x² − 2x + 5) dx = (3/2) ∫ (2x − 2)/(x² − 2x + 5) dx + ∫ dx/(x² − 2x + 5)
First piece = (3/2) log(x² − 2x + 5).
Second: x² − 2x + 5 = (x − 1)² + 2², so it gives (1/2) tan⁻¹((x − 1)/2).
Answer: (3/2) log(x² − 2x + 5) + (1/2) tan⁻¹((x − 1)/2) + C.
Derivative of the denominator is 2x − 2. Set 3x − 2 = λ(2x − 2) + μ.
Comparing x terms: 3 = 2λ → λ = 3/2.
Comparing constants: −2 = −2λ + μ = −3 + μ → μ = 1.
∫ (3x − 2)/(x² − 2x + 5) dx = (3/2) ∫ (2x − 2)/(x² − 2x + 5) dx + ∫ dx/(x² − 2x + 5)
First piece = (3/2) log(x² − 2x + 5).
Second: x² − 2x + 5 = (x − 1)² + 2², so it gives (1/2) tan⁻¹((x − 1)/2).
Answer: (3/2) log(x² − 2x + 5) + (1/2) tan⁻¹((x − 1)/2) + C.
Example 37 — The same trick under a root
Find ∫ (x + 2)/√(x² + 2x + 5) dx.
Derivative of x² + 2x + 5 is 2x + 2. Set x + 2 = λ(2x + 2) + μ:
1 = 2λ → λ = 1/2; 2 = 2λ + μ = 1 + μ → μ = 1.
∫ (x+2)/√(x²+2x+5) dx = (1/2) ∫ (2x+2)/√(x²+2x+5) dx + ∫ dx/√(x²+2x+5)
First piece: t = x² + 2x + 5 gives (1/2) ∫ t⁻¹⁄² dt = (1/2)·2√t = √(x² + 2x + 5).
Second piece: x² + 2x + 5 = (x + 1)² + 2², so it gives log |(x + 1) + √((x+1)² + 4)|.
Answer: √(x² + 2x + 5) + log |(x + 1) + √(x² + 2x + 5)| + C.
Derivative of x² + 2x + 5 is 2x + 2. Set x + 2 = λ(2x + 2) + μ:
1 = 2λ → λ = 1/2; 2 = 2λ + μ = 1 + μ → μ = 1.
∫ (x+2)/√(x²+2x+5) dx = (1/2) ∫ (2x+2)/√(x²+2x+5) dx + ∫ dx/√(x²+2x+5)
First piece: t = x² + 2x + 5 gives (1/2) ∫ t⁻¹⁄² dt = (1/2)·2√t = √(x² + 2x + 5).
Second piece: x² + 2x + 5 = (x + 1)² + 2², so it gives log |(x + 1) + √((x+1)² + 4)|.
Answer: √(x² + 2x + 5) + log |(x + 1) + √(x² + 2x + 5)| + C.
Example 38 — When μ turns out to be zero
Find ∫ (2x + 1)/√(x² + x + 1) dx.
The derivative of x² + x + 1 is exactly 2x + 1 — the numerator already is the derivative, so μ = 0 and there is no second piece at all.
Put t = x² + x + 1, dt = (2x + 1) dx:
∫ dt/√t = 2√t + C
Answer: 2√(x² + x + 1) + C.
Always check for this shortcut before doing the full split — it saves a page of work.
The derivative of x² + x + 1 is exactly 2x + 1 — the numerator already is the derivative, so μ = 0 and there is no second piece at all.
Put t = x² + x + 1, dt = (2x + 1) dx:
∫ dt/√t = 2√t + C
Answer: 2√(x² + x + 1) + C.
Always check for this shortcut before doing the full split — it saves a page of work.
Exam Tip
Write the derivative of the denominator on your rough sheet first, before touching the numerator. Half the marks in this type are for setting the split up correctly, and examiners award them for the λ and μ working even if you slip later.
Integrals Involving Square Roots Of Quadratics
The last three entries on the syllabus list are ∫ √(a² − x²) dx, ∫ √(x² + a²) dx, ∫ √(x² − a²) dx and their general cousin ∫ √(ax² + bx + c) dx. Here the root is on top, not underneath, and the three results share one memorable shape.
Key Rule — all three at once
Each answer is (x/2) × the root, plus (a²/2) × the corresponding “dx over the root” integral:
∫ √(a² − x²) dx = (x/2)√(a² − x²) + (a²/2) sin⁻¹(x/a) + C
∫ √(x² + a²) dx = (x/2)√(x² + a²) + (a²/2) log |x + √(x² + a²)| + C
∫ √(x² − a²) dx = (x/2)√(x² − a²) − (a²/2) log |x + √(x² − a²)| + C
Only the third has a minus in the middle. The second half of each formula is exactly the standard integral of 1 over that same root — which is why learning the previous sections first pays off here.
∫ √(a² − x²) dx = (x/2)√(a² − x²) + (a²/2) sin⁻¹(x/a) + C
∫ √(x² + a²) dx = (x/2)√(x² + a²) + (a²/2) log |x + √(x² + a²)| + C
∫ √(x² − a²) dx = (x/2)√(x² − a²) − (a²/2) log |x + √(x² − a²)| + C
Only the third has a minus in the middle. The second half of each formula is exactly the standard integral of 1 over that same root — which is why learning the previous sections first pays off here.
And for the general quadratic under a root: complete the square, then apply whichever of the three formulas the resulting pattern matches, with (x + b/2) playing the role of x. That is all ∫ √(ax² + bx + c) dx ever is.
Example 39 — The sin⁻¹ version
Find ∫ √(4 − x²) dx.
Here a² = 4, so a = 2, matching √(a² − x²).
∫ √(4 − x²) dx = (x/2)√(4 − x²) + (4/2) sin⁻¹(x/2) + C
Answer: (x/2)√(4 − x²) + 2 sin⁻¹(x/2) + C.
Check: differentiating the first term gives √(4−x²)/2 − x²/(2√(4−x²)); the second gives 2·(1/2)/√(1 − x²/4) = 2/√(4−x²). Adding over a common denominator: [(4 − x²) − x² + 4]/(2√(4−x²)) = (8 − 2x²)/(2√(4−x²)) = √(4 − x²). ✓
Here a² = 4, so a = 2, matching √(a² − x²).
∫ √(4 − x²) dx = (x/2)√(4 − x²) + (4/2) sin⁻¹(x/2) + C
Answer: (x/2)√(4 − x²) + 2 sin⁻¹(x/2) + C.
Check: differentiating the first term gives √(4−x²)/2 − x²/(2√(4−x²)); the second gives 2·(1/2)/√(1 − x²/4) = 2/√(4−x²). Adding over a common denominator: [(4 − x²) − x² + 4]/(2√(4−x²)) = (8 − 2x²)/(2√(4−x²)) = √(4 − x²). ✓
Example 40 — The minus-in-the-middle version
Find ∫ √(x² − 9) dx.
a = 3, and this is the x² − a² form — the one with the minus sign in the middle of the answer.
Answer: (x/2)√(x² − 9) − (9/2) log |x + √(x² − 9)| + C.
If you write a plus there by habit, you lose the mark. Say it out loud when you learn it: “minus a squared over two, only for x squared minus a squared.”
a = 3, and this is the x² − a² form — the one with the minus sign in the middle of the answer.
Answer: (x/2)√(x² − 9) − (9/2) log |x + √(x² − 9)| + C.
If you write a plus there by habit, you lose the mark. Say it out loud when you learn it: “minus a squared over two, only for x squared minus a squared.”
Example 41 — The general quadratic under the root
Find ∫ √(x² + 2x + 5) dx.
Complete the square first:
x² + 2x + 5 = (x + 1)² + 4 = (x + 1)² + 2²
Now use the √(x² + a²) formula with a = 2 and x replaced by (x + 1):
= ((x + 1)/2)√((x+1)² + 4) + (4/2) log |(x + 1) + √((x+1)² + 4)| + C
Answer: ((x + 1)/2)√(x² + 2x + 5) + 2 log |(x + 1) + √(x² + 2x + 5)| + C.
Complete the square first:
x² + 2x + 5 = (x + 1)² + 4 = (x + 1)² + 2²
Now use the √(x² + a²) formula with a = 2 and x replaced by (x + 1):
= ((x + 1)/2)√((x+1)² + 4) + (4/2) log |(x + 1) + √((x+1)² + 4)| + C
Answer: ((x + 1)/2)√(x² + 2x + 5) + 2 log |(x + 1) + √(x² + 2x + 5)| + C.
Example 42 — One more, straight from the table
Find ∫ √(x² + 16) dx.
a = 4, so a²/2 = 8, and this is the plus version (so the middle sign is plus).
Answer: (x/2)√(x² + 16) + 8 log |x + √(x² + 16)| + C.
a = 4, so a²/2 = 8, and this is the plus version (so the middle sign is plus).
Answer: (x/2)√(x² + 16) + 8 log |x + √(x² + 16)| + C.
Exam Tip
Notice the family resemblance running through the last four sections. ∫ dx/(quadratic), ∫ dx/√(quadratic), ∫ (px+q)/(quadratic) and ∫ √(quadratic) dx all begin with the same move: complete the square. If you can do that reliably under pressure, this entire block of the syllabus becomes one technique with four endings.
Choosing The Right Method
You now own every technique in the chapter. The remaining skill — and honestly the one that decides your marks — is looking at a fresh integral and knowing which tool to reach for. Work down this checklist in order. It resolves the vast majority of questions you will meet.
Work down the green questions in order; the first YES tells you the method to use.
Exam Tip — build your own instinct
Take twenty past-paper integrals and, without solving any of them, just write the method next to each one. Ten minutes of that trains the recognition reflex far faster than solving five questions slowly. Method-spotting is a separate skill from method-executing, and it deserves its own practice.
Key Idea — more than one road
Many integrals can be done by two different methods and both are fully correct. ∫ sin x cos x dx yields to substitution (t = sin x), to a trigonometric identity (sin 2x / 2), or to a different substitution (t = cos x) — and the three answers differ only by a constant, which the “+ C” absorbs. If your answer looks unlike the textbook one, differentiate it before assuming you are wrong.
Definite Integrals And The Fundamental Theorem
Everything so far produced a family of functions with a “+ C” hanging off it. A definite integral is different: it produces a single number. We write it with limits — a lower limit a and an upper limit b:
∫ab f(x) dx
Geometrically this number is the signed area between the curve y = f(x) and the x-axis, from x = a to x = b. “Signed” is the important word. Area above the axis counts as positive, and area below the axis counts as negative. Think of it like a bank statement over a month: deposits count up, withdrawals count down, and the definite integral reports your net change, not the total money that moved.
The integral reports the NET of the two areas (−2/3 + 4/3 = 2/3), not the total area swept (which would be 2).
Key Rule — the Fundamental Theorem of Calculus
If f is continuous on [a, b] and F is any anti-derivative of f (that is, F′(x) = f(x)), then
∫ab f(x) dx = [F(x)]ab = F(b) − F(a)
(stated here without proof, as your syllabus specifies). This is one of the great results in mathematics: it says that areas and anti-derivatives — two ideas with no obvious connection — are the same thing. Practically, it means you never have to compute an area from scratch. Integrate, substitute the upper limit, substitute the lower limit, subtract.
Notice you can drop the “+ C”: if you used F(x) + C, the constants cancel in F(b) − F(a). That is why definite integrals never carry a + C.
∫ab f(x) dx = [F(x)]ab = F(b) − F(a)
(stated here without proof, as your syllabus specifies). This is one of the great results in mathematics: it says that areas and anti-derivatives — two ideas with no obvious connection — are the same thing. Practically, it means you never have to compute an area from scratch. Integrate, substitute the upper limit, substitute the lower limit, subtract.
Notice you can drop the “+ C”: if you used F(x) + C, the constants cancel in F(b) − F(a). That is why definite integrals never carry a + C.
Example 43 — Your first definite integral
Evaluate ∫01 x² dx.
Anti-derivative: F(x) = x³/3.
∫01 x² dx = [x³/3]01 = 1³/3 − 0³/3 = 1/3.
Answer: 1/3. A number, not a function — and no + C anywhere.
Anti-derivative: F(x) = x³/3.
∫01 x² dx = [x³/3]01 = 1³/3 − 0³/3 = 1/3.
Answer: 1/3. A number, not a function — and no + C anywhere.
Example 44 — A trigonometric one
Evaluate ∫0π/2 sin x dx.
∫ sin x dx = − cos x, so
[− cos x]0π/2 = (− cos(π/2)) − (− cos 0) = (−0) − (−1) = 1.
Answer: 1.
Notice how carefully the subtraction had to be done — the lower limit brought its own minus sign, and two minuses made a plus. Write the brackets explicitly; do not do this step in your head.
∫ sin x dx = − cos x, so
[− cos x]0π/2 = (− cos(π/2)) − (− cos 0) = (−0) − (−1) = 1.
Answer: 1.
Notice how carefully the subtraction had to be done — the lower limit brought its own minus sign, and two minuses made a plus. Write the brackets explicitly; do not do this step in your head.
Example 45 — When the curve dips below the axis
Evaluate ∫02 |x − 1| dx.
The modulus changes definition at x = 1, so split the interval there:
On [0, 1], x − 1 ≤ 0, so |x − 1| = 1 − x.
On [1, 2], x − 1 ≥ 0, so |x − 1| = x − 1.
∫01 (1 − x) dx = [x − x²/2]01 = 1 − 1/2 = 1/2
∫12 (x − 1) dx = [x²/2 − x]12 = (2 − 2) − (1/2 − 1) = 1/2
Answer: 1/2 + 1/2 = 1.
Whenever a modulus, or a function defined piecewise, appears inside a definite integral, split at the point where the definition changes. You cannot integrate across a break in the formula.
The modulus changes definition at x = 1, so split the interval there:
On [0, 1], x − 1 ≤ 0, so |x − 1| = 1 − x.
On [1, 2], x − 1 ≥ 0, so |x − 1| = x − 1.
∫01 (1 − x) dx = [x − x²/2]01 = 1 − 1/2 = 1/2
∫12 (x − 1) dx = [x²/2 − x]12 = (2 − 2) − (1/2 − 1) = 1/2
Answer: 1/2 + 1/2 = 1.
Whenever a modulus, or a function defined piecewise, appears inside a definite integral, split at the point where the definition changes. You cannot integrate across a break in the formula.
Common Mistake
Substituting the limits in the wrong order. It is always upper minus lower, F(b) − F(a). Reversing them flips the sign of your entire answer, and it is such a quiet error that students rarely spot it on re-reading. Also: never write “+ C” on a definite integral — it signals to the examiner that you have not understood what a definite integral is.
Definite Integrals By Substitution
When a definite integral needs a substitution, you have a choice, and one of the two routes is much safer under exam pressure.
Key Rule — change the limits with the variable
Method A (recommended). Substitute t = g(x), and change the limits too: when x = a, t = g(a); when x = b, t = g(b). Then evaluate entirely in t. You never return to x at all.
Method B. Find the indefinite integral, substitute back to x, and only then apply the original limits.
Method A is faster and eliminates the biggest error in the topic — applying x-limits to a t-expression. Whichever you choose, never mix them.
Method B. Find the indefinite integral, substitute back to x, and only then apply the original limits.
Method A is faster and eliminates the biggest error in the topic — applying x-limits to a t-expression. Whichever you choose, never mix them.
Example 46 — Changing the limits
Evaluate ∫01 x/(1 + x²) dx.
Put t = 1 + x², so dt = 2x dx, i.e. x dx = dt/2.
Change the limits: when x = 0, t = 1. When x = 1, t = 2.
∫01 x/(1 + x²) dx = (1/2) ∫12 dt/t = (1/2)[log t]12 = (1/2)(log 2 − log 1)
Answer: (1/2) log 2 ≈ 0.3466.
Because log 1 = 0, the lower limit contributed nothing — but you must still write it down.
Put t = 1 + x², so dt = 2x dx, i.e. x dx = dt/2.
Change the limits: when x = 0, t = 1. When x = 1, t = 2.
∫01 x/(1 + x²) dx = (1/2) ∫12 dt/t = (1/2)[log t]12 = (1/2)(log 2 − log 1)
Answer: (1/2) log 2 ≈ 0.3466.
Because log 1 = 0, the lower limit contributed nothing — but you must still write it down.
Example 47 — A trigonometric substitution with limits
Evaluate ∫0π/4 tan x sec² x dx.
The derivative of tan x is sec² x, which is sitting right there. Put t = tan x, dt = sec² x dx.
Change the limits: when x = 0, t = tan 0 = 0. When x = π/4, t = tan(π/4) = 1.
∫0π/4 tan x sec² x dx = ∫01 t dt = [t²/2]01 = 1/2
Answer: 1/2.
The derivative of tan x is sec² x, which is sitting right there. Put t = tan x, dt = sec² x dx.
Change the limits: when x = 0, t = tan 0 = 0. When x = π/4, t = tan(π/4) = 1.
∫0π/4 tan x sec² x dx = ∫01 t dt = [t²/2]01 = 1/2
Answer: 1/2.
Example 48 — By parts with limits
Evaluate ∫01 x eˣ dx.
From Example 19, the anti-derivative is (x − 1)eˣ.
[(x − 1)eˣ]01 = (1 − 1)e¹ − (0 − 1)e⁰ = 0 − (−1) = 1
Answer: 1.
Definite integration adds nothing new to the by-parts method — find the anti-derivative exactly as before, then apply the limits at the very end.
From Example 19, the anti-derivative is (x − 1)eˣ.
[(x − 1)eˣ]01 = (1 − 1)e¹ − (0 − 1)e⁰ = 0 − (−1) = 1
Answer: 1.
Definite integration adds nothing new to the by-parts method — find the anti-derivative exactly as before, then apply the limits at the very end.
Common Mistake
The classic disaster: substituting t = 1 + x² and then writing ∫01 dt/t with the old limits still attached. Those limits belong to x, not to t. Either change them (Method A) or convert back to x first (Method B) — but doing neither gives a completely wrong number with working that looks perfectly reasonable.
Properties Of Definite Integrals
Here is where definite integrals stop being routine and start being clever. Some integrals are genuinely impossible to evaluate by finding an anti-derivative — and yet their definite value can be found in three lines using a property. These properties are pure exam gold, and they are explicitly named in your syllabus.
| Property | What it is really saying | |
|---|---|---|
| P₀ | ∫ab f(x) dx = ∫ab f(t) dt | The variable of integration is a dummy — its name does not matter. |
| P₁ | ∫ab f(x) dx = − ∫ba f(x) dx | Swapping the limits flips the sign. In particular ∫aa f(x) dx = 0. |
| P₂ | ∫ab f(x) dx = ∫ac f(x) dx + ∫cb f(x) dx | You may break the interval at any point c. This is what you use for modulus and piecewise functions. |
| P₃ | ∫ab f(x) dx = ∫ab f(a + b − x) dx | Reflect the interval end-to-end. The workhorse for ∫0π and similar. |
| P₄ | ∫0a f(x) dx = ∫0a f(a − x) dx | The special case of P₃ with the lower limit 0. The single most-used property in the exam. |
| P₅ | ∫02a f(x) dx = ∫0a f(x) dx + ∫0a f(2a − x) dx | Splits a double-length interval into two of length a. |
| P₆ | ∫02a f(x) dx = 2 ∫0a f(x) dx if f(2a − x) = f(x); = 0 if f(2a − x) = − f(x) | A symmetry test on [0, 2a]. |
| P₇ | ∫−aa f(x) dx = 2 ∫0a f(x) dx if f is even; = 0 if f is odd | The even/odd property. Check this first on any symmetric interval — it can turn a page of work into a single zero. |
Key Idea — why P₄ is so powerful
P₄ says ∫0a f(x) dx = ∫0a f(a − x) dx. Call the original integral I. Then the reflected version is also equal to I. So you may add the two together:
2I = ∫0a [ f(x) + f(a − x) ] dx
and very often that sum collapses to something trivial — frequently just the constant 1. You then get I in one step, without ever finding an anti-derivative. This “add the integral to its own reflection” move is the single most valuable trick in the chapter.
2I = ∫0a [ f(x) + f(a − x) ] dx
and very often that sum collapses to something trivial — frequently just the constant 1. You then get I in one step, without ever finding an anti-derivative. This “add the integral to its own reflection” move is the single most valuable trick in the chapter.
Exam Tip — the reflex order
Facing any definite integral, run these three checks before doing any work: (1) Are the limits symmetric, −a to a? Then test odd/even (P₇). (2) Does the integrand look ugly but the limits look neat (0 to π/2, 0 to π, 0 to a)? Then try P₄. (3) Is there a modulus or a piecewise definition? Then split with P₂. Only if all three fail should you reach for an anti-derivative.
Putting The Properties To Work
Reading the properties is not enough — you have to watch them bite. Here are the standard patterns, and once you have done these you will start recognising them everywhere.
Example 49 — The odd-function shortcut (P₇)
Evaluate ∫−11 x³ cos x dx.
The limits are symmetric, so test the parity. Let f(x) = x³ cos x. Then
f(−x) = (−x)³ cos(−x) = − x³ cos x = − f(x)
because x³ is odd and cos x is even, and odd × even = odd.
By P₇, the integral of an odd function over a symmetric interval is zero.
Answer: 0.
Finding an anti-derivative of x³ cos x needs three rounds of integration by parts. The property gets you there in two lines. Always check the parity first.
The limits are symmetric, so test the parity. Let f(x) = x³ cos x. Then
f(−x) = (−x)³ cos(−x) = − x³ cos x = − f(x)
because x³ is odd and cos x is even, and odd × even = odd.
By P₇, the integral of an odd function over a symmetric interval is zero.
Answer: 0.
Finding an anti-derivative of x³ cos x needs three rounds of integration by parts. The property gets you there in two lines. Always check the parity first.
Example 50 — The even-function halving (P₇)
Evaluate ∫−22 (x² + 1) dx.
f(−x) = (−x)² + 1 = x² + 1 = f(x), so f is even.
By P₇: ∫−22 (x² + 1) dx = 2 ∫02 (x² + 1) dx = 2[x³/3 + x]02 = 2(8/3 + 2) = 2(14/3)
Answer: 28/3.
Halving the interval halves the chance of an arithmetic slip — the lower limit becomes 0, which usually contributes nothing.
f(−x) = (−x)² + 1 = x² + 1 = f(x), so f is even.
By P₇: ∫−22 (x² + 1) dx = 2 ∫02 (x² + 1) dx = 2[x³/3 + x]02 = 2(8/3 + 2) = 2(14/3)
Answer: 28/3.
Halving the interval halves the chance of an arithmetic slip — the lower limit becomes 0, which usually contributes nothing.
Example 51 — The famous P₄ trick
Evaluate I = ∫0π/2 dx/(1 + tan x).
Try to find an anti-derivative and you will get nowhere. But the limits 0 to π/2 are a loud hint. Apply P₄ with a = π/2, replacing x by (π/2 − x):
tan(π/2 − x) = cot x, so
I = ∫0π/2 dx/(1 + cot x)
Now rewrite that using cot x = cos x/sin x:
1/(1 + cot x) = sin x/(sin x + cos x)
And the original, using tan x = sin x/cos x:
1/(1 + tan x) = cos x/(sin x + cos x)
Add the two expressions for I:
2I = ∫0π/2 [cos x + sin x]/(sin x + cos x) dx = ∫0π/2 1 dx = π/2
Answer: I = π/4 ≈ 0.7854.
Look at what happened: the impossible integrand cancelled to the number 1. That is the whole design of P₄.
Try to find an anti-derivative and you will get nowhere. But the limits 0 to π/2 are a loud hint. Apply P₄ with a = π/2, replacing x by (π/2 − x):
tan(π/2 − x) = cot x, so
I = ∫0π/2 dx/(1 + cot x)
Now rewrite that using cot x = cos x/sin x:
1/(1 + cot x) = sin x/(sin x + cos x)
And the original, using tan x = sin x/cos x:
1/(1 + tan x) = cos x/(sin x + cos x)
Add the two expressions for I:
2I = ∫0π/2 [cos x + sin x]/(sin x + cos x) dx = ∫0π/2 1 dx = π/2
Answer: I = π/4 ≈ 0.7854.
Look at what happened: the impossible integrand cancelled to the number 1. That is the whole design of P₄.
Example 52 — The same trick, different clothes
Evaluate I = ∫0π/2 √(sin x) / [√(sin x) + √(cos x)] dx.
Apply P₄ with a = π/2. Since sin(π/2 − x) = cos x and cos(π/2 − x) = sin x, the two roots simply swap:
I = ∫0π/2 √(cos x) / [√(cos x) + √(sin x)] dx
Add the two forms — the denominators are identical, so the numerators just add:
2I = ∫0π/2 [√(sin x) + √(cos x)]/[√(sin x) + √(cos x)] dx = ∫0π/2 1 dx = π/2
Answer: I = π/4.
The same answer as Example 51, and for the same structural reason. Swap √ for any power, or sin³ for √sin, and the answer is still π/4. Recognise the shape, not the particular function.
Apply P₄ with a = π/2. Since sin(π/2 − x) = cos x and cos(π/2 − x) = sin x, the two roots simply swap:
I = ∫0π/2 √(cos x) / [√(cos x) + √(sin x)] dx
Add the two forms — the denominators are identical, so the numerators just add:
2I = ∫0π/2 [√(sin x) + √(cos x)]/[√(sin x) + √(cos x)] dx = ∫0π/2 1 dx = π/2
Answer: I = π/4.
The same answer as Example 51, and for the same structural reason. Swap √ for any power, or sin³ for √sin, and the answer is still π/4. Recognise the shape, not the particular function.
Example 53 — P₄ when the reflection does not cancel
Evaluate I = ∫0π x sin x/(1 + cos² x) dx.
Apply P₄ with a = π. Note sin(π − x) = sin x and cos(π − x) = − cos x, so cos²(π − x) = cos² x. Only the lone x changes:
I = ∫0π (π − x) sin x/(1 + cos² x) dx
Add the two expressions:
2I = ∫0π [x + (π − x)] sin x/(1 + cos² x) dx = π ∫0π sin x/(1 + cos² x) dx
The awkward x has vanished. Now substitute t = cos x, dt = − sin x dx. Limits: x = 0 → t = 1; x = π → t = −1.
∫0π sin x/(1 + cos² x) dx = − ∫1−1 dt/(1 + t²) = ∫−11 dt/(1 + t²) = [tan⁻¹t]−11 = π/4 − (−π/4) = π/2
So 2I = π · (π/2) = π²/2.
Answer: I = π²/4 ≈ 2.4674.
This is the model answer for a whole family of board questions. The pattern to memorise: when a stray x multiplies a function that is symmetric about the midpoint of the interval, P₄ removes the x.
Apply P₄ with a = π. Note sin(π − x) = sin x and cos(π − x) = − cos x, so cos²(π − x) = cos² x. Only the lone x changes:
I = ∫0π (π − x) sin x/(1 + cos² x) dx
Add the two expressions:
2I = ∫0π [x + (π − x)] sin x/(1 + cos² x) dx = π ∫0π sin x/(1 + cos² x) dx
The awkward x has vanished. Now substitute t = cos x, dt = − sin x dx. Limits: x = 0 → t = 1; x = π → t = −1.
∫0π sin x/(1 + cos² x) dx = − ∫1−1 dt/(1 + t²) = ∫−11 dt/(1 + t²) = [tan⁻¹t]−11 = π/4 − (−π/4) = π/2
So 2I = π · (π/2) = π²/2.
Answer: I = π²/4 ≈ 2.4674.
This is the model answer for a whole family of board questions. The pattern to memorise: when a stray x multiplies a function that is symmetric about the midpoint of the interval, P₄ removes the x.
Common Mistake
When you apply P₄ you must replace every x in the integrand by (a − x) — including the x sitting outside the fraction, as in Example 53. Changing only the trigonometric parts and leaving a bare x behind is the mistake that quietly destroys these questions. Also remember that the limits do not change under P₄; only the integrand does.
Common Mistakes And How To Avoid Them
Almost every mark lost in this chapter comes from one of the following. Read this list the night before your exam — it is the cheapest revision you will ever do.
The forgotten + C
An indefinite integral without “+ C” is an incomplete answer, and boards do deduct for it. Build a habit: the moment you write the integral sign, write “+ C” at the far right of the line. Fill in the middle afterwards.
The missing modulus in a log
∫ dx/x = log |x| + C, not log x + C. The same applies to log |sec x|, log |x − a| and every other log that comes out of an integration. Drop the modulus only when the argument is provably positive, such as log(1 + x²).
Inventing a product rule
There is no rule that says ∫ f g dx = (∫ f)(∫ g). Products need by parts or substitution. Quotients need partial fractions or a rewrite. If you catch yourself multiplying two separate integrals together, stop.
Losing the constant in a substitution
If t = x² then dt = 2x dx, so x dx = dt/2 — that 1/2 must appear in your answer. Immediately after writing your substitution, write the dx-replacement on the same line, before touching the integrand. Differentiate your final answer to confirm.
Using x-limits on a t-integral
After substituting in a definite integral, the old limits no longer apply. Either change them to t-limits or convert back to x before substituting. Mixing the two produces a confident-looking wrong answer.
Sign slips in the special formulas
Three specific ones cost marks every year: ∫ sin x dx = − cos x (not + cos x); ∫ √(x² − a²) dx has a minus before the (a²/2) log term while the other two have a plus; and ∫ (x + 1)⁻² dx = − 1/(x + 1), where the power rule flips the sign.
Guessing the partial-fraction shape
A linear factor gets a constant on top; an irreducible quadratic gets a linear expression (Bx + C); a factor repeated k times needs k separate terms. And check the fraction is proper before you start — if it is not, divide first.
Ignoring the properties
If the limits are −a to a, or 0 to a with an ugly integrand, and you launch straight into finding an anti-derivative, you are almost certainly doing five times the necessary work. Spend ten seconds on the parity and P₄ checks before committing.
Good to Know — the one habit that fixes most of this
Differentiate your answer. Every time. It takes twenty seconds, it needs no extra knowledge, and it catches the missing constant, the wrong sign, the dropped 1/2 and the misapplied formula. No other single habit in Class 12 Maths returns as many marks per second spent.
Practice Worksheet
Ten original questions, arranged from warm-up to board level. Cover the answers, attempt each one on paper, and only then reveal. A question you got wrong and then understood is worth more than three you got right.
Q1. Find ∫ (3x² − 4/x² + 5) dx.
Show Answer
Integrate term by term. ∫3x² dx = x³. For the middle term write 4/x² = 4x⁻², so ∫(−4x⁻²) dx = −4 · x⁻¹/(−1) = +4/x. And ∫5 dx = 5x.
Answer: x³ + 4/x + 5x + C.
Check: d/dx = 3x² − 4/x² + 5. ✓ Note how the minus sign became a plus — that is the power rule with a negative index.
Answer: x³ + 4/x + 5x + C.
Check: d/dx = 3x² − 4/x² + 5. ✓ Note how the minus sign became a plus — that is the power rule with a negative index.
Q2. Find ∫ x²/(1 + x³) dx.
Show Answer
The derivative of 1 + x³ is 3x², and x² is on top — so this is a substitution.
Put t = 1 + x³, dt = 3x² dx, so x² dx = dt/3.
∫ x²/(1 + x³) dx = (1/3) ∫ dt/t = (1/3) log |t| + C
Answer: (1/3) log |1 + x³| + C.
Put t = 1 + x³, dt = 3x² dx, so x² dx = dt/3.
∫ x²/(1 + x³) dx = (1/3) ∫ dt/t = (1/3) log |t| + C
Answer: (1/3) log |1 + x³| + C.
Q3. Find ∫ cos² x dx.
Show Answer
Use the power-reducing identity cos² x = (1 + cos 2x)/2.
∫ cos² x dx = (1/2) ∫ (1 + cos 2x) dx = (1/2)[x + (sin 2x)/2] + C
Answer: x/2 + (sin 2x)/4 + C.
Check: d/dx = 1/2 + (cos 2x)/2 = (1 + cos 2x)/2 = cos² x. ✓
∫ cos² x dx = (1/2) ∫ (1 + cos 2x) dx = (1/2)[x + (sin 2x)/2] + C
Answer: x/2 + (sin 2x)/4 + C.
Check: d/dx = 1/2 + (cos 2x)/2 = (1 + cos 2x)/2 = cos² x. ✓
Q4. Find ∫ (2x + 5)/((x + 1)(x + 4)) dx.
Show Answer
Partial fractions, Case 1. Assume (2x + 5)/((x+1)(x+4)) = A/(x+1) + B/(x+4).
Multiply up: 2x + 5 = A(x + 4) + B(x + 1).
Put x = −1: 3 = A(3) → A = 1.
Put x = −4: −3 = B(−3) → B = 1.
∫ [1/(x+1) + 1/(x+4)] dx = log |x + 1| + log |x + 4| + C
Answer: log |(x + 1)(x + 4)| + C.
A pleasant surprise: both constants came out as 1.
Multiply up: 2x + 5 = A(x + 4) + B(x + 1).
Put x = −1: 3 = A(3) → A = 1.
Put x = −4: −3 = B(−3) → B = 1.
∫ [1/(x+1) + 1/(x+4)] dx = log |x + 1| + log |x + 4| + C
Answer: log |(x + 1)(x + 4)| + C.
A pleasant surprise: both constants came out as 1.
Q5. Find ∫ x cos x dx.
Show Answer
A product of an algebraic and a trigonometric function, so integration by parts. By LIATE, A beats T, so u = x and the second function is cos x.
∫ x cos x dx = x sin x − ∫ (1)(sin x) dx = x sin x − (− cos x) + C
Answer: x sin x + cos x + C.
Check: d/dx = sin x + x cos x − sin x = x cos x. ✓ Watch the double negative in the second term — it is where most people slip.
∫ x cos x dx = x sin x − ∫ (1)(sin x) dx = x sin x − (− cos x) + C
Answer: x sin x + cos x + C.
Check: d/dx = sin x + x cos x − sin x = x cos x. ✓ Watch the double negative in the second term — it is where most people slip.
Q6. Find ∫ eˣ (tan x + sec² x) dx.
Show Answer
Test the eˣ[f + f′] pattern. Is sec² x the derivative of tan x? Yes, exactly.
So f(x) = tan x, f′(x) = sec² x, and the answer is eˣ f(x).
Answer: eˣ tan x + C.
Check: d/dx[eˣ tan x] = eˣ tan x + eˣ sec² x = eˣ(tan x + sec² x). ✓ No working needed — only recognition.
So f(x) = tan x, f′(x) = sec² x, and the answer is eˣ f(x).
Answer: eˣ tan x + C.
Check: d/dx[eˣ tan x] = eˣ tan x + eˣ sec² x = eˣ(tan x + sec² x). ✓ No working needed — only recognition.
Q7. Find ∫ dx/(x² − 2x + 10).
Show Answer
Complete the square. Half of −2 is −1, and (−1)² = 1:
x² − 2x + 10 = (x² − 2x + 1) + 9 = (x − 1)² + 3²
This is the tan⁻¹ row with a = 3 and x replaced by (x − 1).
Answer: (1/3) tan⁻¹((x − 1)/3) + C.
x² − 2x + 10 = (x² − 2x + 1) + 9 = (x − 1)² + 3²
This is the tan⁻¹ row with a = 3 and x replaced by (x − 1).
Answer: (1/3) tan⁻¹((x − 1)/3) + C.
Q8. Find ∫ (x + 3)/√(x² + 6x + 13) dx.
Show Answer
Split the numerator using the derivative of the quadratic, which is 2x + 6.
Set x + 3 = λ(2x + 6) + μ. Comparing x terms: 1 = 2λ → λ = 1/2. Comparing constants: 3 = 6λ + μ = 3 + μ → μ = 0.
So there is no second piece at all — the numerator is exactly half the derivative.
Put t = x² + 6x + 13, dt = (2x + 6) dx:
(1/2) ∫ dt/√t = (1/2)(2√t) = √t
Answer: √(x² + 6x + 13) + C.
Check: d/dx√(x²+6x+13) = (2x + 6)/(2√(x²+6x+13)) = (x + 3)/√(x²+6x+13). ✓
Set x + 3 = λ(2x + 6) + μ. Comparing x terms: 1 = 2λ → λ = 1/2. Comparing constants: 3 = 6λ + μ = 3 + μ → μ = 0.
So there is no second piece at all — the numerator is exactly half the derivative.
Put t = x² + 6x + 13, dt = (2x + 6) dx:
(1/2) ∫ dt/√t = (1/2)(2√t) = √t
Answer: √(x² + 6x + 13) + C.
Check: d/dx√(x²+6x+13) = (2x + 6)/(2√(x²+6x+13)) = (x + 3)/√(x²+6x+13). ✓
Q9. Find ∫ √(9 − x²) dx.
Show Answer
This is the ∫√(a² − x²) dx formula with a² = 9, so a = 3 and a²/2 = 9/2.
∫ √(a² − x²) dx = (x/2)√(a² − x²) + (a²/2) sin⁻¹(x/a) + C
Answer: (x/2)√(9 − x²) + (9/2) sin⁻¹(x/3) + C.
Remember the middle sign is plus here; only the √(x² − a²) version takes a minus.
∫ √(a² − x²) dx = (x/2)√(a² − x²) + (a²/2) sin⁻¹(x/a) + C
Answer: (x/2)√(9 − x²) + (9/2) sin⁻¹(x/3) + C.
Remember the middle sign is plus here; only the √(x² − a²) version takes a minus.
Q10. Evaluate ∫−π/2π/2 x⁵ sin² x dx.
Show Answer
Stop before integrating — the limits are symmetric, so check the parity. Let f(x) = x⁵ sin² x.
f(−x) = (−x)⁵ sin²(−x) = −x⁵ · (−sin x)² = −x⁵ sin² x = − f(x)
So f is odd (odd × even = odd), and by P₇ the integral over a symmetric interval is zero.
Answer: 0.
Finding an anti-derivative here would take several pages. The parity check takes fifteen seconds.
f(−x) = (−x)⁵ sin²(−x) = −x⁵ · (−sin x)² = −x⁵ sin² x = − f(x)
So f is odd (odd × even = odd), and by P₇ the integral over a symmetric interval is zero.
Answer: 0.
Finding an anti-derivative here would take several pages. The parity check takes fifteen seconds.
Q11. Evaluate ∫0π/2 cos x/(1 + sin² x) dx.
Show Answer
A function and its derivative: the derivative of sin x is cos x, which is on top.
Put t = sin x, dt = cos x dx. Change the limits: x = 0 → t = 0; x = π/2 → t = 1.
∫01 dt/(1 + t²) = [tan⁻¹t]01 = tan⁻¹1 − tan⁻¹0 = π/4 − 0
Answer: π/4 ≈ 0.7854.
Put t = sin x, dt = cos x dx. Change the limits: x = 0 → t = 0; x = π/2 → t = 1.
∫01 dt/(1 + t²) = [tan⁻¹t]01 = tan⁻¹1 − tan⁻¹0 = π/4 − 0
Answer: π/4 ≈ 0.7854.
Q12. Evaluate I = ∫0π/2 sin³x/(sin³x + cos³x) dx.
Show Answer
Ugly integrand, neat limits → reach for P₄.
Apply ∫0a f(x) dx = ∫0a f(a − x) dx with a = π/2. Since sin(π/2 − x) = cos x and cos(π/2 − x) = sin x, the cubes swap places:
I = ∫0π/2 cos³x/(cos³x + sin³x) dx
Add the two expressions for I. The denominators are the same, so the numerators combine:
2I = ∫0π/2 (sin³x + cos³x)/(sin³x + cos³x) dx = ∫0π/2 1 dx = π/2
Answer: I = π/4.
Exactly the structure of Examples 51 and 52 — once you own this shape, a whole family of board questions becomes routine.
Apply ∫0a f(x) dx = ∫0a f(a − x) dx with a = π/2. Since sin(π/2 − x) = cos x and cos(π/2 − x) = sin x, the cubes swap places:
I = ∫0π/2 cos³x/(cos³x + sin³x) dx
Add the two expressions for I. The denominators are the same, so the numerators combine:
2I = ∫0π/2 (sin³x + cos³x)/(sin³x + cos³x) dx = ∫0π/2 1 dx = π/2
Answer: I = π/4.
Exactly the structure of Examples 51 and 52 — once you own this shape, a whole family of board questions becomes routine.
Kaizen, not perfection. You do not need to master this chapter today. Aim for one more correct question than yesterday — just one. Do that consistently and by the time the board exam arrives, integration will be the part of the paper you look forward to.