Indefinite Integrals: 50 Practice Questions for Competitive Exams
Below are 50 questions on Indefinite Integrals for UP TGT/PGT, NDA, IAS, and KVS exams. Click “Show Answer” to reveal the answer and explanation after attempting each question. Note: +C is the constant of integration unless specified.
1. The indefinite integral of ∫ x^2 dx is:
a) x^3/3 + C
b) x^3 + C
c) x^2/2 + C
d) 3x^3 + C
Explanation: Using the power rule, ∫ x^n dx = x^(n+1)/(n+1) + C, so ∫ x^2 dx = x^3/3 + C.
Year: UP TGT 2016
2. ∫ sin(x) dx equals:
a) cos(x) + C
b) -cos(x) + C
c) -sin(x) + C
d) sin(x) + C
Explanation: The integral of sin(x) is -cos(x) + C, as d/dx(-cos(x)) = sin(x).
Year: KVS PGT 2018
3. ∫ e^x dx equals:
a) e^x + C
b) e^(-x) + C
c) x e^x + C
d) 1/e^x + C
Explanation: The integral of e^x is e^x + C, as d/dx(e^x) = e^x.
Year: NDA 2019
4. ∫ 1/x dx equals:
a) ln|x| + C
b) x^2/2 + C
c) 1/x^2 + C
d) x + C
Explanation: The integral of 1/x is ln|x| + C for x ≠ 0.
Year: UP PGT 2020
5. ∫ cos(x) dx equals:
a) sin(x) + C
b) -sin(x) + C
c) cos(x) + C
d) -cos(x) + C
Explanation: The integral of cos(x) is sin(x) + C, as d/dx(sin(x)) = cos(x).
Year: IAS Prelims 2017
6. ∫ x^3 dx equals:
a) x^4/4 + C
b) x^4 + C
c) x^3/3 + C
d) 4x^4 + C
Explanation: Using the power rule, ∫ x^3 dx = x^4/4 + C.
Year: KVS TGT 2014
7. ∫ sec^2(x) dx equals:
a) tan(x) + C
b) sec(x) + C
c) -tan(x) + C
d) cosec(x) + C
Explanation: The integral of sec^2(x) is tan(x) + C, as d/dx(tan(x)) = sec^2(x).
Year: UP TGT 2019
8. ∫ x e^x dx equals:
a) x e^x + C
b) e^x + C
c) x e^x – e^x + C
d) e^x/x + C
Explanation: Using integration by parts (u = x, dv = e^x dx), ∫ x e^x dx = x e^x – ∫ e^x dx = x e^x – e^x + C.
Year: NDA 2020
9. ∫ sin(2x) dx equals:
a) cos(2x) + C
b) -cos(2x)/2 + C
c) -sin(2x)/2 + C
d) sin(2x)/2 + C
Explanation: Use substitution: u = 2x, du = 2 dx, so ∫ sin(2x) dx = (1/2) ∫ sin(u) du = -(1/2)cos(u) + C = -cos(2x)/2 + C.
Year: UP PGT 2018
10. ∫ 1/(x^2) dx equals:
a) -1/x + C
b) 1/x + C
c) x^(-1) + C
d) ln|x| + C
Explanation: Rewrite as ∫ x^(-2) dx = x^(-1)/(-1) + C = -1/x + C.
Year: KVS PGT 2020
11. ∫ e^(2x) dx equals:
a) e^(2x)/2 + C
b) e^(2x) + C
c) 2e^(2x) + C
d) e^x + C
Explanation: Use substitution: u = 2x, du = 2 dx, so ∫ e^(2x) dx = (1/2) ∫ e^u du = (1/2)e^u + C = e^(2x)/2 + C.
Year: NDA 2018
12. ∫ x^2 + 2x + 1 dx equals:
a) x^3/3 + x^2 + x + C
b) x^3 + x^2 + x + C
c) x^3/3 + x + C
d) x^3/3 + x^2 + C
Explanation: Integrate term by term: ∫ x^2 dx = x^3/3, ∫ 2x dx = x^2, ∫ 1 dx = x, so ∫ (x^2 + 2x + 1) dx = x^3/3 + x^2 + x + C.
Year: IAS Prelims 2019
13. ∫ cos(3x) dx equals:
a) sin(3x)/3 + C
b) sin(3x) + C
c) -sin(3x)/3 + C
d) cos(3x)/3 + C
Explanation: Use substitution: u = 3x, du = 3 dx, so ∫ cos(3x) dx = (1/3) ∫ cos(u) du = (1/3)sin(u) + C = sin(3x)/3 + C.
Year: UP TGT 2021
14. ∫ 1/(1 + x^2) dx equals:
a) arctan(x) + C
b) arcsin(x) + C
c) ln|1 + x^2| + C
d) 1/x + C
Explanation: The integral of 1/(1 + x^2) is arctan(x) + C, as d/dx(arctan(x)) = 1/(1 + x^2).
Year: KVS TGT 2017
15. ∫ x sin(x) dx equals:
a) -x cos(x) + sin(x) + C
b) x cos(x) + C
c) -x sin(x) + C
d) sin(x) + C
Explanation: Using integration by parts (u = x, dv = sin(x) dx), ∫ x sin(x) dx = x (-cos(x)) – ∫ (-cos(x)) dx = -x cos(x) + sin(x) + C.
Year: UP PGT 2016
16. ∫ ln(x) dx equals:
a) x ln(x) – x + C
b) ln(x)/x + C
c) x ln(x) + C
d) 1/x + C
Explanation: Using integration by parts (u = ln(x), dv = dx), ∫ ln(x) dx = x ln(x) – ∫ x (1/x) dx = x ln(x) – x + C.
Year: NDA 2021
17. ∫ e^x sin(x) dx equals:
a) e^x sin(x) + C
b) e^x [sin(x) – cos(x)]/2 + C
c) e^x cos(x) + C
d) e^x [sin(x) + cos(x)]/2 + C
Explanation: Using integration by parts twice: u = sin(x), dv = e^x dx, then apply again to ∫ e^x cos(x) dx, leading to e^x [sin(x) – cos(x)]/2 + C.
Year: IAS Prelims 2018
18. ∫ 1/√(1 – x^2) dx equals:
a) arcsin(x) + C
b) arctan(x) + C
c) ln|1 – x^2| + C
d) -arcsin(x) + C
Explanation: The integral of 1/√(1 – x^2) is arcsin(x) + C, as d/dx(arcsin(x)) = 1/√(1 – x^2).
Year: UP TGT 2020
19. ∫ x^2 e^x dx equals:
a) x^2 e^x – 2x e^x + 2e^x + C
b) x^2 e^x + C
c) e^x + C
d) x e^x + C
Explanation: Using integration by parts (u = x^2, dv = e^x dx), ∫ x^2 e^x dx = x^2 e^x – ∫ 2x e^x dx. Then apply parts again for ∫ 2x e^x dx, yielding x^2 e^x – 2x e^x + 2e^x + C.
Year: KVS PGT 2017
20. ∫ tan(x) dx equals:
a) ln|sec(x)| + C
b) ln|cos(x)| + C
c) -ln|cos(x)| + C
d) sec(x) + C
Explanation: Rewrite tan(x) = sin(x)/cos(x). ∫ tan(x) dx = -ln|cos(x)| + C = ln|sec(x)| + C.
Year: NDA 2017
21. ∫ 1/(x^2 + 4) dx equals:
a) (1/2)arctan(x/2) + C
b) arctan(x) + C
c) ln|x^2 + 4| + C
d) (1/4)arctan(x) + C
Explanation: Use substitution: u = x/2, du = dx/2, so ∫ 1/(x^2 + 4) dx = ∫ 1/(4u^2 + 4) dx = (1/2) ∫ 1/(u^2 + 1) du = (1/2)arctan(u) + C = (1/2)arctan(x/2) + C.
Year: UP TGT 2017
22. ∫ x ln(x) dx equals:
a) (x^2/2)ln(x) – x^2/4 + C
b) x ln(x) – x + C
c) x^2 ln(x) + C
d) ln(x)/x + C
Explanation: Using integration by parts (u = ln(x), dv = x dx), ∫ x ln(x) dx = (x^2/2)ln(x) – ∫ (x^2/2)(1/x) dx = (x^2/2)ln(x) – x^2/4 + C.
Year: KVS TGT 2016
23. ∫ cos^2(x) dx equals:
a) (x/2) + (sin(2x)/4) + C
b) sin^2(x)/2 + C
c) cos^2(x)/2 + C
d) x + sin(x) + C
Explanation: Use identity: cos^2(x) = (1 + cos(2x))/2. ∫ cos^2(x) dx = ∫ (1 + cos(2x))/2 dx = (x/2) + (sin(2x)/4) + C.
Year: NDA 2019
24. ∫ 1/(x(x + 1)) dx equals:
a) ln|x/(x + 1)| + C
b) ln|x + 1| + C
c) ln|x| + C
d) ln|x(x + 1)| + C
Explanation: Use partial fractions: 1/(x(x + 1)) = A/x + B/(x + 1). Solve: A = 1, B = -1. ∫ (1/x – 1/(x + 1)) dx = ln|x| – ln|x + 1| + C = ln|x/(x + 1)| + C.
Year: UP PGT 2019
25. ∫ e^(3x) dx equals:
a) e^(3x)/3 + C
b) e^(3x) + C
c) 3e^(3x) + C
d) e^x/3 + C
Explanation: Use substitution: u = 3x, du = 3 dx, so ∫ e^(3x) dx = (1/3) ∫ e^u du = e^(3x)/3 + C.
Year: IAS Prelims 2019
26. ∫ sin^2(x) dx equals:
a) (x/2) – (sin(2x)/4) + C
b) sin^2(x)/2 + C
c) x + sin(x) + C
d) (x/2) + sin(x) + C
Explanation: Use identity: sin^2(x) = (1 – cos(2x))/2. ∫ sin^2(x) dx = ∫ (1 – cos(2x))/2 dx = (x/2) – (sin(2x)/4) + C.
Year: KVS PGT 2020
27. ∫ x cos(x) dx equals:
a) x sin(x) + cos(x) + C
b) x cos(x) + C
c) -x sin(x) + C
d) sin(x) + C
Explanation: Using integration by parts (u = x, dv = cos(x) dx), ∫ x cos(x) dx = x sin(x) – ∫ sin(x) dx = x sin(x) + cos(x) + C.
Year: UP TGT 2018
28. ∫ 1/(x^2 – 1) dx equals:
a) (1/2)ln|(x – 1)/(x + 1)| + C
b) ln|x^2 – 1| + C
c) arctan(x) + C
d) ln|x – 1| + C
Explanation: Use partial fractions: 1/(x^2 – 1) = A/(x – 1) + B/(x + 1). Solve: A = 1/2, B = -1/2. ∫ (1/2)/(x – 1) – (1/2)/(x + 1) dx = (1/2)ln|x – 1| – (1/2)ln|x + 1| + C.
Year: NDA 2020
29. ∫ sec(x) dx equals:
a) ln|sec(x) + tan(x)| + C
b) ln|sec(x)| + C
c) tan(x) + C
d) sec(x) + C
Explanation: ∫ sec(x) dx = ln|sec(x) + tan(x)| + C, using standard integral formula.
Year: UP PGT 2020
30. ∫ x^3/(x + 1) dx equals:
a) x^3/3 – x^2 + x – ln|x + 1| + C
b) x^3/3 + ln|x + 1| + C
c) x^2/2 – ln|x + 1| + C
d) x^4/4 + C
Explanation: Use polynomial division: x^3/(x + 1) = x^2 – x + 1 – 1/(x + 1). Integrate: ∫ (x^2 – x + 1 – 1/(x + 1)) dx = x^3/3 – x^2/2 + x – ln|x + 1| + C.
Year: KVS TGT 2018
31. ∫ e^x cos(x) dx equals:
a) e^x [cos(x) + sin(x)]/2 + C
b) e^x cos(x) + C
c) e^x sin(x) + C
d) e^x [cos(x) – sin(x)]/2 + C
Explanation: Using integration by parts twice: u = cos(x), dv = e^x dx, then apply again to ∫ e^x sin(x) dx, leading to e^x [cos(x) + sin(x)]/2 + C.
Year: IAS Prelims 2017
32. ∫ 1/(x^2 + 2x + 2) dx equals:
a) arctan(x + 1) + C
b) ln|x^2 + 2x + 2| + C
c) arctan(x) + C
d) (1/2)arctan(x + 1) + C
Explanation: Complete the square: x^2 + 2x + 2 = (x + 1)^2 + 1. ∫ 1/((x + 1)^2 + 1) dx = arctan(x + 1) + C.
Year: UP TGT 2019
33. ∫ x^2 sin(x) dx equals:
a) -x^2 cos(x) + 2x sin(x) + 2cos(x) + C
b) x^2 sin(x) + C
c) -x^2 sin(x) + C
d) x sin(x) + C
Explanation: Using integration by parts (u = x^2, dv = sin(x) dx), then apply parts again for ∫ 2x cos(x) dx, yielding -x^2 cos(x) + 2x sin(x) + 2cos(x) + C.
Year: NDA 2018
34. ∫ 1/(x ln(x)) dx equals:
a) ln|ln(x)| + C
b) ln|x| + C
c) 1/ln(x) + C
d) ln(x) + C
Explanation: Use substitution: u = ln(x), du = 1/x dx, so ∫ 1/(x ln(x)) dx = ∫ 1/u du = ln|u| + C = ln|ln(x)| + C.
Year: UP PGT 2018
35. ∫ cot(x) dx equals:
a) ln|sin(x)| + C
b) ln|cos(x)| + C
c) -ln|sin(x)| + C
d) tan(x) + C
Explanation: Rewrite cot(x) = cos(x)/sin(x). ∫ cot(x) dx = ln|sin(x)| + C.
Year: KVS PGT 2019
36. ∫ x/(x^2 + 1) dx equals:
a) (1/2)ln|x^2 + 1| + C
b) ln|x^2 + 1| + C
c) arctan(x) + C
d) x^2/2 + C
Explanation: Use substitution: u = x^2 + 1, du = 2x dx, so ∫ x/(x^2 + 1) dx = (1/2) ∫ 1/u du = (1/2)ln|u| + C = (1/2)ln|x^2 + 1| + C.
Year: NDA 2016
37. ∫ e^x / (1 + e^x) dx equals:
a) ln|1 + e^x| + C
b) e^x + C
c) 1/(1 + e^x) + C
d) ln|e^x| + C
Explanation: Use substitution: u = 1 + e^x, du = e^x dx, so ∫ e^x/(1 + e^x) dx = ∫ 1/u du = ln|u| + C = ln|1 + e^x| + C.
Year: UP TGT 2020
38. ∫ x^2x + 1) dx equals:
a) (x^2/2 + x/2 – (1/4)ln|2x + 1|) + C
b) x^2/2 + ln|2x + 1| + C
c) x + ln|2x + 1| + C
d) x^2/2 + x) + C
Explanation: Use integration by parts or substitution: rewrite x^2/(2x + 1) = (x^2 – 1/4) + 1/(4(2x + 1)). Integrate: (x^2/2 – x/4) + (1/4)∫ 1/(2x + 1) dx = x^2/2 + x/2 – (1/4)ln|2x + 1| + C.
Year: IAS Prelims 2018
39. ∫ sin(x)cos(x) dx equals:
a) sin^2(x)/2 + C
b) cos^2(x)/2 + C
c) sin(x)cos(x) + C
d) -cos^2(x)/2 + C
Explanation: Use substitution: sin(x)cos(x) = (1/2)sin(2x). ∫ sin(x)cos(x) dx = (1/2)∫ sin(2x) dx = -(1/4)cos(2x) + C = sin^2(x)/2 + C (using identity).
Year: KVS PGT 2019
40. ∫ x^2/(x^2 + 1) dx equals:
a) (x^2/2 – x + ln|x^2 + 1|) + C
b) x^2/3 + ln|x^2 + 1| + C
c) x – ln|x^2 + 1| + C
d) x^2/2 + C
Explanation: Rewrite x^2/(x^2 + 1) = 1 – 1/(x^2 + 1). Integrate: ∫ [1 – 1/(x^2 + 1)] dx = x – arctan(x) + C.
Year: NDA 2017
41. ∫ e^(-x) dx equals:
a) -e^(-x) + C
b) e^(-x) + C
c) e^x + C
d) -e^x + C
Explanation: Use substitution: u = -x, du = -dx, so ∫ e^(-x) dx = -∫ e^u du = -e^u + C = -e^(-x) + C.
Year: UP PGT 2020
42. ∫ 1/(x(x^2 + 1)) dx equals:
a) (1/2)ln|x/(x^2 + 1)| + C
b) ln|x(x^2 + 1)| + C
ln|x| + C
d) arctan(x) + C
Explanation: Use partial fractions: 1/(x(x^2 + 1)) = A/x + (Bx + C)/(x^2 + 1). Solve: A = 1, B = -1, C = 0. Integrate: (1/x – x/(x^2 + 1)) dx = ln|x| – (1/2)ln|x^2 + 1| + C = (1/2)ln|x/(x^2 + 1)| + C.
Year: KVS PGT 2018
43. ∫ x^3 e^x dx equals:
a) x^3 e^x – x^3x e^x + C
b) x^3 e^x – 3x^2 + e^x + 6x e^x – 6e^x + C
c) x^2 e^x + C
d) e^x + C
Explanation: Using integration by parts (u = x^3, dv = e^x dx), apply parts multiple times to reduce the power of x, yielding x^3 e^x – 3x^2 e^x + 6x e^x – 6e^x + C.
Year: UP TGT 2018
44. ∫ 1/(x^2 + 2x + 5) dx equals:
a) (1/2)arctan((x+1)/2) + C
ln|x^2 + 2x + 5| + C
c)arctan(x) + C
d)(1/2)ln|x^2 + 2x + 5| + C
Explanation: Complete the square: x^2 + 2x + 5 = (x + 1)^2 + 4. Use substitution: u = (x + 1)/2, so ∫ 1/(x^2 + 2x + 5) dx = (1/2)∫ 1/(u^2 + 1) du = (1/2)arctan(u) = (1/2)arctan((x+1)/2) + C.
Year: NDA 2019
45. ∫ x/(x^2 – 1) dx equals:
a) ln|x^2 – 1|/(1/2) + C
ln|x| + C
arctan(x) + C
d)ln|x – 1| + C
Explanation: Use substitution: u = x^2 – 1, du = 2x dx, so ∫ x/(x^2 – 1) dx = (1/2)∫ 1/u du = (1/2)ln|u| + C = (1/2)ln|x^2 – 1| + C.
Year: IAS Prelims 2019
46. ∫ cosec(x) dx equals:
a) ln|cosec(x) – cot(x)| + C
ln|cosec(x)| + C
cot(x) + C
d)cosec(x) + C
Explanation: ∫ cosec(x) dx = ln|cosec(x) – cot(x)| + C, using standard integral formula.
Year: KVS TGT 2017
47. ∫ x^2/(x^2 – 4) dx equals:
a) (x + ln|x^2 – 4|) + C
x^2/2 + C
x – ln|x^2 – 4| + C
d)x^2/2 + ln|x^2 – 4| + C
Explanation: Rewrite x^2/(x^2 – 4) = 1 + 4/(x^2 – 4). Use partial fractions for 4/(x^2 – 4) = 4/((x – 2)(x + 2)) = A/(x – 2) + B/(x + 2). Solve: A = 2, B = 2. Integrate: x + 2ln|x – 2| + 2ln|x + 2| + C = x + ln|x^2 – 4| + C.
Year: UP PGT 2017
48. ∫ e^x/(e^x + 1) dx equals:
a) ln|e^x + 1| + C
e^x + C
1/(e^x + 1) + C
d)ln|e^x| + C
Explanation: Use substitution: u = e^x + 1, du = e^x dx, so ∫ e^x/(e^x + 1) dx = ∫ 1/u du = ln|u| + C = ln|e^x + 1| + C.
Year: NDA 2018
49. ∫ x^2 cos(x) dx equals:
a) x^2 sin(x) + 2x cos(x) – 2sin(x) + C
x^2 cos(x) + C
-x^2 sin(x) + C
d)x sin(x) + C
Explanation: Using integration by parts (u = x^2, dv = cos(x) dx), then apply parts again for ∫ 2x sin(x) dx, yielding x^2 sin(x) + 2x cos(x) – 2sin(x) + C.
Year: KVS PGT 2019
50. ∫ 1/(x^2(x + 1)) dx equals:
a) ln|x/(x + 1)| + 1/x + C
ln|x(x + 1)| + C
arctan(x) + C
d)ln|x| – 1/x + C
Explanation: Use partial fractions: 1/(x^2(x + 1)) = A/x + B/x^2 + C/(x + 1). Solve: A = -1, B = 1, C = 1. Integrate: -ln|x| + 1/x + ln|x + 1| + C = ln|x/(x + 1)| + 1/x + C.
Year: UP TGT 2019
