Take the ECAT Mathematics — Integration (Set 6): 30 multiple-choice questions (40 minutes) with instant scoring and a full answer review. Covers 30 ECAT…
The first 5 questions of this test, with the correct answer and explanation.
Q1. ∫ x/(x⁴ + 1) dx equals:
- A. tan⁻¹(x²) + C
- B. ½ tan⁻¹(x²) + C ✓
- C. ½ ln(x⁴ + 1) + C
- D. ¼ tan⁻¹(x⁴) + C
Answer: B. ½ tan⁻¹(x²) + C
Explanation: Put u = x²: du = 2x dx, so the integral is ½∫du/(1 + u²) = ½ tan⁻¹(x²) + C.
Q2. ∫ sin x cos x/(1 + sin⁴x) dx equals:
- A. tan⁻¹(sin²x) + C
- B. ½ tan⁻¹(sin x) + C
- C. ½ tan⁻¹(sin²x) + C ✓
- D. ¼ ln(1 + sin⁴x) + C
Answer: C. ½ tan⁻¹(sin²x) + C
Explanation: Let u = sin²x, du = 2 sin x cos x dx. Then ∫ = ½∫du/(1 + u²) = ½ tan⁻¹(sin²x) + C.
Q3. ∫ eˣ(x − 1)/x² dx equals:
- A. eˣ ln x + C
- B. −eˣ/x + C
- C. eˣ/x + C ✓
- D. eˣ(x − 1)/x + C
Answer: C. eˣ/x + C
Explanation: Write eˣ(x − 1)/x² = eˣ[1/x − 1/x²], which has the form eˣ[f(x) + f′(x)] with f = 1/x. Hence ∫ = eˣ/x + C.
Q4. ∫ (x² + 1)/(x⁴ + 1) dx equals:
- A. (1/√2) tan⁻¹[(x² − 1)/(√2 x)] + C ✓
- B. (1/√2) tan⁻¹(x²) + C
- C. (1/2) tan⁻¹[(x² − 1)/x] + C
- D. (1/√2) ln[(x² − 1)/(√2 x)] + C
Answer: A. (1/√2) tan⁻¹[(x² − 1)/(√2 x)] + C
Explanation: Divide numerator and denominator by x²: ∫(1 + 1/x²)dx/(x² + 1/x²). Let u = x − 1/x, du = (1 + 1/x²)dx and x² + 1/x² = u² + 2. So ∫du/(u² + 2) = (1/√2)tan⁻¹(u/√2) = (1/√2)tan⁻¹[(x² − 1)/(√2 x)] + C.
Q5. ∫ sin 3x cos x dx equals:
- A. −cos 4x/4 − cos 2x/2 + C
- B. −cos 4x/8 − cos 2x/4 + C ✓
- C. −cos 3x sin x/3 + C
- D. cos 4x/8 + cos 2x/4 + C
Answer: B. −cos 4x/8 − cos 2x/4 + C
Explanation: sin 3x cos x = ½(sin 4x + sin 2x). Integrating: ½(−cos 4x/4 − cos 2x/2) = −cos 4x/8 − cos 2x/4 + C.
Take the timed test (free, no sign-up) to see your score and the answer and explanation for every question.
Q6. ∫ tan³x sec x dx equals:
- A. tan⁴x/4 + C
- B. tan⁴x/4 · sec x + C
- C. sec³x/3 − sec x + C
- D. sec³x/3 + sec x + C
Q7. ∫ x√(x + 2) dx equals:
- A. (2/5)(x + 2)^(5/2) + (4/3)(x + 2)^(3/2) + C
- B. (2/5)(x + 2)^(5/2) − (2/3)(x + 2)^(3/2) + C
- C. (x²/2)·(2/3)(x + 2)^(3/2) + C
- D. (2/5)(x + 2)^(5/2) − (4/3)(x + 2)^(3/2) + C
Q8. ∫ dx/√(x² + 4x + 13) equals:
- A. sin⁻¹((x + 2)/3) + C
- B. ln|x + 2 + √(x² + 4x + 13)| + C
- C. 2√(x² + 4x + 13) + C
- D. tan⁻¹((x + 2)/3) + C
Q9. ∫ dx/[x(x² + 1)] equals:
- A. ln|x| − tan⁻¹x + C
- B. ½ ln|x| − ln(x² + 1) + C
- C. ln|x| − ½ ln(x² + 1) + C
- D. ln|x| + ½ ln(x² + 1) + C
Q10. ∫ sin²x cos²x dx equals:
- A. x/4 − sin 4x/16 + C
- B. sin³x cos³x/9 + C
- C. x/8 − sin 4x/32 + C
- D. x/8 + sin 4x/32 + C
Q11. ∫ dx/(1 + tan x) equals:
- A. ½[x + ln|sin x + cos x|] + C
- B. ln|1 + tan x| + C
- C. x + ln|sin x + cos x| + C
- D. ½[x − ln|sin x + cos x|] + C
Q12. ∫ x/(x + 1)² dx equals:
- A. ln|x + 1| − 1/(x + 1) + C
- B. x²/[2(x + 1)] + C
- C. ln|x + 1| + 1/(x + 1) + C
- D. ln|x + 1| + 1/(x + 1)² + C
Q13. ∫ sec²x/(tan²x + 4) dx equals:
- A. tan⁻¹(tan x/2) + C
- B. ½ tan⁻¹(tan x/2) + C
- C. ½ tan⁻¹(tan x) + C
- D. ¼ tan⁻¹(tan x/4) + C
Q14. ∫ eˣ/(e²ˣ − 1) dx equals:
- A. tan⁻¹(eˣ) + C
- B. ln|e²ˣ − 1| + C
- C. ½ ln|(eˣ − 1)/(eˣ + 1)| + C
- D. ½ ln|(eˣ + 1)/(eˣ − 1)| + C
Q15. For x > 1, ∫ dx/(x√(x² − 1)) equals:
- A. −sec⁻¹x + C
- B. sec⁻¹x + C
- C. √(x² − 1) + C
- D. sin⁻¹(1/x) + C
Q16. A particle has acceleration a = 6t − 4 m/s², starts at s = 0 with v = 3 m/s at t = 0. Its position at t = 2 s is:
- A. 10 m
- B. 6 m
- C. 4 m
- D. 8 m
Q17. For which value of k does the curve with dy/dx = kx − 6 have its minimum point at x = 3?
- A. k = 6
- B. k = 2
- C. k = 18
- D. k = 3
Q18. ∫ x²/(x⁶ + 4) dx equals:
- A. (1/6) ln(x⁶ + 4) + C
- B. (1/6) tan⁻¹(x³) + C
- C. (1/6) tan⁻¹(x³/2) + C
- D. (1/3) tan⁻¹(x³/2) + C
Q19. ∫ dx/[x√(1 − (ln x)²)] equals:
- A. cos⁻¹(ln x) + C
- B. sin⁻¹(x) · ln x + C
- C. ln|ln x| + C
- D. sin⁻¹(ln x) + C
Q20. ∫ sin x/(1 + cos x)² dx equals:
- A. −1/(1 + cos x) + C
- B. ln(1 + cos x) + C
- C. 1/(1 + cos x) + C
- D. 1/(1 + cos x)² + C
Q21. ∫ x³/√(1 + x²) dx equals:
- A. x⁴/[4√(1 + x²)] + C
- B. (1 + x²)^(3/2)/3 − √(1 + x²) + C
- C. (1 + x²)^(3/2)/6 − √(1 + x²) + C
- D. (1 + x²)^(3/2)/3 + √(1 + x²) + C
Q22. ∫ cos(ln x) dx equals:
- A. (x/2)[cos(ln x) + sin(ln x)] + C
- B. (x/2)[cos(ln x) − sin(ln x)] + C
- C. x[cos(ln x) + sin(ln x)] + C
- D. sin(ln x)/x + C
Q23. ∫ x cosec²x dx equals:
- A. −x cot x − ln|sin x| + C
- B. −x²cot x/2 + C
- C. x cot x + ln|sin x| + C
- D. −x cot x + ln|sin x| + C
Q24. ∫ eˣ(1 + tan x + tan²x) dx equals:
- A. eˣ(1 + tan x) + C
- B. eˣ tan²x + C
- C. eˣ tan x + C
- D. eˣ sec x + C
Q25. ∫ eˣ/√(e²ˣ + 4) dx equals:
- A. ln[eˣ + √(e²ˣ + 4)] + C
- B. 2 sinh⁻¹(eˣ/2) + C
- C. √(e²ˣ + 4) + C
- D. tan⁻¹(eˣ/2)/2 + C
Q26. ∫ dx/[x(x⁵ + 1)] equals:
- A. (1/5) ln(x⁵ + 1) + C
- B. ln[x⁵/(x⁵ + 1)] + C
- C. (1/5) ln[(x⁵ + 1)/x⁵] + C
- D. (1/5) ln[x⁵/(x⁵ + 1)] + C
Q27. ∫ tan⁻¹(√x) dx equals:
- A. (x − 1) tan⁻¹√x − √x + C
- B. (x + 1) tan⁻¹√x + √x + C
- C. x tan⁻¹√x − √x + C
- D. (x + 1) tan⁻¹√x − √x + C
Q28. ∫ dx/(sin²x cos²x) equals:
- A. tan x − cot x + C
- B. tan x + cot x + C
- C. sec x · cosec x + C
- D. cot x − tan x + C
Q29. ∫ (2x + 3)/(x² + 2x + 5) dx equals:
- A. ln(x² + 2x + 5) + ½ tan⁻¹(x + 1) + C
- B. ½ ln(x² + 2x + 5) + tan⁻¹[(x + 1)/2] + C
- C. ln(x² + 2x + 5) + tan⁻¹[(x + 1)/2] + C
- D. ln(x² + 2x + 5) + ½ tan⁻¹[(x + 1)/2] + C
Q30. A curve satisfies d²y/dx² = 6x − 4. At x = 1 its gradient is 3, and it passes through (0, 2). Its y-coordinate at x = 2 is: