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CBSE · Class 12 · Mathematics

Integrals

Standard difficulty · 20 questions · 40 marks

Name: Date: Score: ____ / 40

Show your working for Sections B and C. Answers are on the last page — try every question first!

Section A — Choose the correct answer8 × 1 = 8 marks
1.
∫ (−3x2 + 4x + 2) dx equals:
  1. −x3 + 2x2 + 2x + C
  2. −6x + 4 + C
  3. −3x3 + 4x2 + 2x + C
  4. −x3 + 2x2 − 2x + C
2.
∫ sec29x dx equals:
  1. 9 tan 9x + C
  2. (1/9) sec 9x tan 9x + C
  3. tan 9x + C
  4. (1/9) tan 9x + C
3.
∫−21 (3x + 1) dx equals:
  1. −3/2
  2. 5/2
  3. 13/2
  4. 3/2
4.
∫−11 (−4x3 + x2 + 3x − 5) dx equals:
  1. −28/3
  2. −14/3
  3. 0
  4. −25/6
5.
∫ dx/x equals:
  1. log|x| + C
  2. −1/x2 + C
  3. x + C
  4. 1/x2 + C
6.
∫ x cos x dx equals:
  1. x sin x + cos x + C
  2. x sin x + C
  3. x sin x − cos x + C
  4. −x sin x + cos x + C
7.
∫ dx/((x + 1)(x + 6)) equals:
  1. (1/5) log|(x + 1)/(x + 6)| + C
  2. (1/5) log|(x + 6)/(x + 1)| + C
  3. −(1/5) log|(x + 1)/(x + 6)| + C
  4. log|(x + 1)(x + 6)| + C
8.
Assertion (A): ∫−11 (4x2 + 5) dx = 0.
Reason (R): If f is an odd function, ∫−11 f(x) dx = 0.
  1. Both A and R are true, and R is the correct explanation of A.
  2. Both A and R are true, but R is not the correct explanation of A.
  3. A is true, but R is false.
  4. A is false, but R is true.
Section B — Short answer8 × 2 = 16 marks
9.
Find ∫ (9x5 + 5/x + 8ex) dx.
10.
Find ∫ x(x2 + 4)2 dx.
11.
Evaluate ∫13 (6x2 + 4x − 5) dx.
12.
Find ∫ 1/(x2 − 16) dx.
13.
Find ∫ xe6x dx.
14.
Find ∫ (5x − 2)/((x − 5)(x − 2)) dx.
15.
Evaluate ∫06 |x − 4| dx.
16.
Find ∫ cos26x dx.
Section C — Word problems4 × 4 = 16 marks
17.
Find ∫ dx/((x − 3)(x − 5)(x − 6)).
18.
Evaluate ∫513 √x/(√x + √(18 − x)) dx.
19.
Find ∫ (x2 − 1)e−4x dx.
20.
Evaluate ∫02 x(x2 + 1) dx.
AmbifulWorksheets
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CBSE · Class 12 · Mathematics

Answer key

Integrals · AW-12M-INT-2026-V1 · For parents — fold back or tear off before handing the sheet to your child.

QAnswerWorking
1(a) −x3 + 2x2 + 2x + Craise each power by one and divide by the new power
2(d) (1/9) tan 9x + Ca standard integral
3(a) −3/2[(3/2)x2 + x] from −2 to 1
4(a) −28/3the odd terms integrate to 0; 2∫01 (x2 − 5) dx
5(a) log|x| + C—
6(a) x sin x + cos x + Cintegrate by parts with u = x
7(a) (1/5) log|(x + 1)/(x + 6)| + C1/((x + 1)(x + 6)) = (1/5)[1/(x + 1) − 1/(x + 6)]
8(d) A is false, but R is true.A is false: 4x2 + 5 is even, so the integral is 2∫01 (4x2 + 5) dx = 38/3. R is true.
9(3/2)x6 + 5 log|x| + 8ex + C∫ xⁿ dx = xⁿ+1/(n + 1), ∫ dx/x = log|x|, ∫ ex dx = ex
10(1/6)(x2 + 4)3 + Cput t = x2 + 4, dt = 2x dx
1158[2x3 + 2x2 − 5x] from 1 to 3
12(1/8) log|(x − 4)/(x + 4)| + Cuse the standard form
13e6x((1/6)x − 1/36) + Cby parts: ∫ u dv = uv − ∫ v du, with u = x
14(23/3) log|x − 5| − (8/3) log|x − 2| + C(5x − 2)/((x − 5)(x − 2)) = A/(x − 5) + B/(x − 2) with A = 23/3, B = −8/3
1510split at x = 4: 8 + 2
16(1/2)x + (1/24) sin 12x + Ccos26x = (1 + cos 12x)/2
17(1/6) log|x − 3| − (1/2) log|x − 5| + (1/3) log|x − 6| + Cpartial fractions: A = 1/6, B = −1/2, C = 1/3 (cover-up rule)
184replace x by 18 − x and add the two forms: 2I = ∫513 1 dx = 8
19e−4x(−(1/4)x2 − (1/8)x + 7/32) + Cintegrate by parts twice, with u = x2 and then u = x
206put t = x2 + 1: (1/2)∫15 t dt = [t2/4] from 1 to 5