Indefinite Integration Practice

Take timed practice tests on Indefinite Integration for JEE Main and JEE Advanced with session-wise drills, score review, and explanation-led revision.

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Indefinite Integration Practice Tests

Six 20-question timed sessions plus one 60-question module test. Each question is original and calibrated from the uploaded material pattern without copying PDF wording.

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JEE Main Session 1

4Questions

60s/QTimer

+4 / -1Scoring

Preview all 15 questions in Indefinite Integration Practice Tests (no login required)

Basic formulas
1. ∫xⁿ dx (n ≠ −1) equals:
- A. xⁿ⁺¹/(n+1) + C (Correct)
- B. nxⁿ⁻¹ + C
- C. xⁿ/n + C
- D. xⁿ⁺¹ + C
Explanation: Power rule for integration: ∫xⁿ dx = xⁿ⁺¹/(n+1) + C, valid for n ≠ −1. For n = −1: ∫1/x dx = ln|x| + C.
Basic formulas
2. ∫eˣ dx equals:
- A. eˣ + C (Correct)
- B. eˣ/x + C
- C. xeˣ + C
- D. e^(x+1) + C
Explanation: The exponential function eˣ is its own integral: ∫eˣ dx = eˣ + C.
Trigonometric integrals
3. ∫sin x dx equals:
- A. −cos x + C (Correct)
- B. cos x + C
- C. −sin x + C
- D. tan x + C
Explanation: ∫sin x dx = −cos x + C. Verify: d/dx(−cos x) = sin x ✓.
Substitution
4. ∫2x dx using power rule gives:
- A. x² + C (Correct)
- B. 2x²/2 + C = x² + C
- C. 2 + C
- D. x²/2 + C
Explanation: ∫2x dx = 2·x²/2 + C = x² + C. Or directly: d/dx(x²) = 2x ✓.
Integration by substitution
5. ∫(2x)(x²+1)⁵ dx equals:
- A. (x²+1)⁶/6 + C (Correct)
- B. (x²+1)⁶ + C
- C. 2(x²+1)⁶/6 + C
- D. 5(x²+1)⁴ + C
Explanation: Let u = x²+1, du = 2x dx. Integral becomes ∫u⁵ du = u⁶/6 + C = (x²+1)⁶/6 + C.
Partial fractions
6. ∫1/(x²−1) dx using partial fractions gives:
- A. (1/2)ln|(x−1)/(x+1)| + C (Correct)
- B. ln|x²−1| + C
- C. 1/(x²−1) + C
- D. (1/2)ln|x+1|/(x−1)| + C
Explanation: 1/(x²−1) = 1/((x−1)(x+1)) = (1/2)[1/(x−1) − 1/(x+1)]. Integrating: (1/2)[ln|x−1| − ln|x+1|] = (1/2)ln|(x−1)/(x+1)| + C.
Integration by parts
7. ∫x eˣ dx equals:
- A. eˣ(x−1) + C (Correct)
- B. xeˣ + C
- C. eˣ/x + C
- D. x²eˣ/2 + C
Explanation: By parts: ∫x eˣ dx = x·eˣ − ∫eˣ dx = xeˣ − eˣ + C = eˣ(x−1) + C. (LIATE: take x as first, eˣ as second.)
Standard integrals
8. ∫1/√(1−x²) dx equals:
- A. sin⁻¹x + C (Correct)
- B. cos⁻¹x + C
- C. tan⁻¹x + C
- D. sec⁻¹x + C
Explanation: Standard form: ∫1/√(1−x²) dx = sin⁻¹x + C (or equivalently −cos⁻¹x + C, differing by a constant).
Reduction formula
9. ∫ln x dx equals:
- A. x ln x − x + C (Correct)
- B. ln x + C
- C. x/ln x + C
- D. ln x / x + C
Explanation: By parts: u = ln x, dv = dx → du = 1/x dx, v = x. ∫ln x dx = x ln x − ∫x·(1/x) dx = x ln x − ∫1 dx = x ln x − x + C.
JEE Main
10. ∫tan x dx equals:
- A. ln|sec x| + C
- B. −ln|cos x| + C
- C. Both A and B are correct (Correct)
- D. sec²x + C
Explanation: ∫tan x dx = ∫sin x / cos x dx. Let u = cos x, du = −sin x dx: −∫du/u = −ln|cos x| + C = ln|sec x| + C. Both forms are equivalent.
Integration by parts twice
11. ∫eˣ sin x dx equals:
- A. eˣ(sin x − cos x)/2 + C (Correct)
- B. eˣ sin x + C
- C. eˣ cos x + C
- D. −eˣ cos x + C
Explanation: Let I = ∫eˣ sin x dx. By parts twice: I = eˣ sin x − ∫eˣ cos x dx = eˣ sin x − (eˣ cos x + ∫eˣ sin x dx) = eˣ(sin x − cos x) − I. 2I = eˣ(sin x − cos x) + C → I = eˣ(sin x − cos x)/2 + C.
JEE Advanced
12. ∫√(tan x) dx is of the form:
- A. Requires splitting into ∫√(tan x)(1+tan²x)/... with substitution tan x = t² (Correct)
- B. Simple power rule gives answer
- C. ∫tan^(1/2) x dx = (2/3)tan^(3/2) x + C
- D. ln|tan x| + C
Explanation: ∫√(tan x) dx is a non-trivial integral. Standard approach: let tan x = t² → sec²x dx = 2t dt. Then ∫t·2t/(1+t⁴) dt = 2∫t²/(1+t⁴) dt, which is solved by decomposing 1+t⁴ = (t²+√2t+1)(t²−√2t+1). This is a JEE Advanced-level technique.
Mixed
13. ∫cos x dx equals:
- A. sin x + C (Correct)
- B. −sin x + C
- C. tan x + C
- D. −cos x + C
Explanation: ∫cos x dx = sin x + C. Verify: d/dx(sin x) = cos x ✓.
Mixed
14. ∫1/(1+x²) dx equals:
- A. tan⁻¹x + C (Correct)
- B. sin⁻¹x + C
- C. sec⁻¹x + C
- D. ln|1+x²| + C
Explanation: Standard form: ∫1/(1+x²) dx = tan⁻¹x + C. Verify: d/dx(tan⁻¹x) = 1/(1+x²) ✓.
Mixed
15. ∫x/(x²+1) dx equals:
- A. (1/2)ln(x²+1) + C (Correct)
- B. ln(x²+1) + C
- C. tan⁻¹x + C
- D. 1/(x²+1) + C
Explanation: Let u = x²+1, du = 2x dx → ∫x/(x²+1) dx = (1/2)∫du/u = (1/2)ln|u| + C = (1/2)ln(x²+1) + C.

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JEE Main Session 1

4Questions

60s/QTimer

+4 / -1Scoring

Preview all 15 questions in Indefinite Integration Practice Tests (no login required)

Basic formulas
1. ∫xⁿ dx (n ≠ −1) equals:
- A. xⁿ⁺¹/(n+1) + C (Correct)
- B. nxⁿ⁻¹ + C
- C. xⁿ/n + C
- D. xⁿ⁺¹ + C
Explanation: Power rule for integration: ∫xⁿ dx = xⁿ⁺¹/(n+1) + C, valid for n ≠ −1. For n = −1: ∫1/x dx = ln|x| + C.
Basic formulas
2. ∫eˣ dx equals:
- A. eˣ + C (Correct)
- B. eˣ/x + C
- C. xeˣ + C
- D. e^(x+1) + C
Explanation: The exponential function eˣ is its own integral: ∫eˣ dx = eˣ + C.
Trigonometric integrals
3. ∫sin x dx equals:
- A. −cos x + C (Correct)
- B. cos x + C
- C. −sin x + C
- D. tan x + C
Explanation: ∫sin x dx = −cos x + C. Verify: d/dx(−cos x) = sin x ✓.
Substitution
4. ∫2x dx using power rule gives:
- A. x² + C (Correct)
- B. 2x²/2 + C = x² + C
- C. 2 + C
- D. x²/2 + C
Explanation: ∫2x dx = 2·x²/2 + C = x² + C. Or directly: d/dx(x²) = 2x ✓.
Integration by substitution
5. ∫(2x)(x²+1)⁵ dx equals:
- A. (x²+1)⁶/6 + C (Correct)
- B. (x²+1)⁶ + C
- C. 2(x²+1)⁶/6 + C
- D. 5(x²+1)⁴ + C
Explanation: Let u = x²+1, du = 2x dx. Integral becomes ∫u⁵ du = u⁶/6 + C = (x²+1)⁶/6 + C.
Partial fractions
6. ∫1/(x²−1) dx using partial fractions gives:
- A. (1/2)ln|(x−1)/(x+1)| + C (Correct)
- B. ln|x²−1| + C
- C. 1/(x²−1) + C
- D. (1/2)ln|x+1|/(x−1)| + C
Explanation: 1/(x²−1) = 1/((x−1)(x+1)) = (1/2)[1/(x−1) − 1/(x+1)]. Integrating: (1/2)[ln|x−1| − ln|x+1|] = (1/2)ln|(x−1)/(x+1)| + C.
Integration by parts
7. ∫x eˣ dx equals:
- A. eˣ(x−1) + C (Correct)
- B. xeˣ + C
- C. eˣ/x + C
- D. x²eˣ/2 + C
Explanation: By parts: ∫x eˣ dx = x·eˣ − ∫eˣ dx = xeˣ − eˣ + C = eˣ(x−1) + C. (LIATE: take x as first, eˣ as second.)
Standard integrals
8. ∫1/√(1−x²) dx equals:
- A. sin⁻¹x + C (Correct)
- B. cos⁻¹x + C
- C. tan⁻¹x + C
- D. sec⁻¹x + C
Explanation: Standard form: ∫1/√(1−x²) dx = sin⁻¹x + C (or equivalently −cos⁻¹x + C, differing by a constant).
Reduction formula
9. ∫ln x dx equals:
- A. x ln x − x + C (Correct)
- B. ln x + C
- C. x/ln x + C
- D. ln x / x + C
Explanation: By parts: u = ln x, dv = dx → du = 1/x dx, v = x. ∫ln x dx = x ln x − ∫x·(1/x) dx = x ln x − ∫1 dx = x ln x − x + C.
JEE Main
10. ∫tan x dx equals:
- A. ln|sec x| + C
- B. −ln|cos x| + C
- C. Both A and B are correct (Correct)
- D. sec²x + C
Explanation: ∫tan x dx = ∫sin x / cos x dx. Let u = cos x, du = −sin x dx: −∫du/u = −ln|cos x| + C = ln|sec x| + C. Both forms are equivalent.
Integration by parts twice
11. ∫eˣ sin x dx equals:
- A. eˣ(sin x − cos x)/2 + C (Correct)
- B. eˣ sin x + C
- C. eˣ cos x + C
- D. −eˣ cos x + C
Explanation: Let I = ∫eˣ sin x dx. By parts twice: I = eˣ sin x − ∫eˣ cos x dx = eˣ sin x − (eˣ cos x + ∫eˣ sin x dx) = eˣ(sin x − cos x) − I. 2I = eˣ(sin x − cos x) + C → I = eˣ(sin x − cos x)/2 + C.
JEE Advanced
12. ∫√(tan x) dx is of the form:
- A. Requires splitting into ∫√(tan x)(1+tan²x)/... with substitution tan x = t² (Correct)
- B. Simple power rule gives answer
- C. ∫tan^(1/2) x dx = (2/3)tan^(3/2) x + C
- D. ln|tan x| + C
Explanation: ∫√(tan x) dx is a non-trivial integral. Standard approach: let tan x = t² → sec²x dx = 2t dt. Then ∫t·2t/(1+t⁴) dt = 2∫t²/(1+t⁴) dt, which is solved by decomposing 1+t⁴ = (t²+√2t+1)(t²−√2t+1). This is a JEE Advanced-level technique.
Mixed
13. ∫cos x dx equals:
- A. sin x + C (Correct)
- B. −sin x + C
- C. tan x + C
- D. −cos x + C
Explanation: ∫cos x dx = sin x + C. Verify: d/dx(sin x) = cos x ✓.
Mixed
14. ∫1/(1+x²) dx equals:
- A. tan⁻¹x + C (Correct)
- B. sin⁻¹x + C
- C. sec⁻¹x + C
- D. ln|1+x²| + C
Explanation: Standard form: ∫1/(1+x²) dx = tan⁻¹x + C. Verify: d/dx(tan⁻¹x) = 1/(1+x²) ✓.
Mixed
15. ∫x/(x²+1) dx equals:
- A. (1/2)ln(x²+1) + C (Correct)
- B. ln(x²+1) + C
- C. tan⁻¹x + C
- D. 1/(x²+1) + C
Explanation: Let u = x²+1, du = 2x dx → ∫x/(x²+1) dx = (1/2)∫du/u = (1/2)ln|u| + C = (1/2)ln(x²+1) + C.