PMA Long Course Mathematics: Integration MCQs

PMA Long Course Mathematics: Integration MCQs — sample questions

1. Question 1

   Q1. ∫(2x + 1) / (x² + x + 1) dx

   • A) ln|x² + x + 1| + C
   • B) ln|x + 1| + C
   • C) (1/2)ln|x² + x + 1| + C
   • D) (1/2)ln|x + 1| + C

   Answer: ln|x² + x + 1| + C

   Explanation: Using substitution u = x² + x + 1, du/dx = 2x + 1, hence ∫(2x + 1) / (x² + x + 1) dx = ∫du/u = ln|u| + C.

2. Question 2

   Q2. ∫(sin x + cos x) / √(sin 2x) dx

   • A) √(sin x + cos x) + C
   • B) √(sin 2x) + C
   • C) sin^-1(sin x - cos x) + C
   • D) ln|sin x + cos x| + C

   Answer: sin^-1(sin x - cos x) + C

   Explanation: Using trigonometric identity sin 2x = (sin x + cos x)² - 1, and substitution, we simplify the integral to ∫1 / √(1 - (sin x - cos x)²) d(sin x - cos x).

3. Question 3

   Q3. ∫1 / (x² - 4) dx on 1 / 2 to 1

   • A) (1/4)ln|3|
   • B) (1/4)ln|(3/5)|
   • C) (1/4)ln|(5/3)|
   • D) -(1/4)ln|(3/5)|

   Answer: (1/4)ln|(3/5)|

   Explanation: Using partial fractions, 1 / (x² - 4) = 1/4 * (1/(x-2) - 1/(x+2)), and then integrating within the given limits.

4. Question 4

   Q4. ∫x² / √(x² + 4) dx

   • A) (x/2)√(x² + 4) - 2ln|x + √(x² + 4)| + C
   • B) (x/2)√(x² + 4) + 2ln|x + √(x² + 4)| + C
   • C) (x/4)√(x² + 4) + 2ln|x + √(x² + 4)| + C
   • D) (x/4)√(x² + 4) - 2ln|x + √(x² + 4)| + C

   Answer: (x/2)√(x² + 4) - 2ln|x + √(x² + 4)| + C

   Explanation: Using trigonometric substitution x = 2tan(θ), dx = 2sec²(θ)dθ, and simplifying.

5. Question 5

   Q5. ∫e^x (1 + x) / (1 + x)² dx

   • A) e^x / (1 + x) + C
   • B) e^x(1 + x) + C
   • C) e^x + C
   • D) e^x(1 - x) + C

   Answer: e^x / (1 + x) + C

   Explanation: Simplifying the integrand to e^x / (1 + x), then using substitution u = 1 + x.

6. Question 6

   Q6. ∫1 / (1 + 3sin²x) dx

   • A) (1/2)tan^-1(2tan x) + C
   • B) (1/√2)tan^-1(√2tan x) + C
   • C) (1/2)tan^-1((1/2)tan x) + C
   • D) (1/√3)tan^-1(√3tan x) + C

   Answer: (1/√3)tan^-1(√3tan x) + C

   Explanation: Dividing numerator and denominator by cos²x, and then using substitution u = tan x.

7. Question 7

   Q7. ∫(x + 1) / √(x² + 2x + 2) dx

   • A) √(x² + 2x + 2) + C
   • B) √(x² + 2x + 2) + ln|x + 1 + √(x² + 2x + 2)| + C
   • C) ln|x + 1 + √(x² + 2x + 2)| + C
   • D) √(x² + 2x + 2) - ln|x + 1 + √(x² + 2x + 2)| + C

   Answer: √(x² + 2x + 2) + ln|x + 1 + √(x² + 2x + 2)| + C

   Explanation: Splitting the integral into two parts, one for √(x² + 2x + 2) and the other for 1 / √(x² + 2x + 2).

8. Question 8

   Q8. ∫sin³x cos x dx

   • A) (1/4)sin⁴x + C
   • B) -(1/4)sin⁴x + C
   • C) (1/4)cos⁴x + C
   • D) -(1/4)cos⁴x + C

   Answer: (1/4)sin⁴x + C

   Explanation: Using substitution u = sin x, du = cos x dx, hence ∫u³ du = (1/4)u⁴ + C.

9. Question 9

   Q9. ∫x / (x + 1) dx on 0 to 1

   • A) 1 - ln 2
   • B) 1 + ln 2
   • C) ln 2
   • D) -ln 2

   Answer: 1 - ln 2

   Explanation: Splitting the integrand into 1 - 1/(x+1), then integrating within the given limits.

10. Question 10

    Q10. ∫1 / √(4 - x²) dx on 0 to 2

    • A) π / 2
    • B) π
    • C) π / 4
    • D) π / 3

    Answer: π / 2

    Explanation: Using trigonometric substitution x = 2sin(θ), dx = 2cos(θ)dθ, and simplifying.

11. Question 11

    Q11. ∫x²e^x dx

    • A) x²e^x - 2xe^x + 2e^x + C
    • B) x²e^x + 2xe^x + 2e^x + C
    • C) x²e^x - 2xe^x - 2e^x + C
    • D) x²e^x + 2xe^x - 2e^x + C

    Answer: x²e^x - 2xe^x + 2e^x + C

    Explanation: Using integration by parts twice, with u = x² and dv = e^x dx.

12. Question 12

    Q12. ∫1 / (x + √(x² + 1)) dx

    • A) ln|x + √(x² + 1)| + C
    • B) (1/2)ln|x + √(x² + 1)| + C
    • C) (1/2)x√(x² + 1) + (1/2)ln|x + √(x² + 1)| + C
    • D) x√(x² + 1) + ln|x + √(x² + 1)| + C

    Answer: (1/2)ln|x + √(x² + 1)| + C

    Explanation: Using substitution x = sinh(u), dx = cosh(u)du, and simplifying.

13. Question 13

    Q13. ∫(sin x + cos x) dx on 0 to π / 2

    • A) 2
    • B) 1
    • C) 0
    • D) -1

    Answer: 2

    Explanation: Integrating sin x + cos x within the given limits, resulting in [-cos x + sin x] from 0 to π / 2.

14. Question 14

    Q14. ∫x / √(x² + 1) dx

    • A) √(x² + 1) + C
    • B) √(x² + 1) / 2 + C
    • C) (1/2)√(x² + 1) + C
    • D) (1/2)x√(x² + 1) + C

    Answer: √(x² + 1) + C

    Explanation: Using substitution u = x² + 1, du = 2x dx, hence (1/2)∫1 / √u du = √u + C.

15. Question 15

    Q15. ∫e^(√x) / √x dx

    • A) 2e^(√x) + C
    • B) e^(√x) + C
    • C) (1/2)e^(√x) + C
    • D) -(1/2)e^(√x) + C

    Answer: 2e^(√x) + C

    Explanation: Using substitution u = √x, du = 1 / (2√x) dx, hence 2∫e^u du = 2e^u + C.

16. Question 16

    Q16. ∫1 / (1 + cos x) dx

    • A) tan(x / 2) + C
    • B) -cot(x / 2) + C
    • C) -tan(x / 2) + C
    • D) cot(x / 2) + C

    Answer: tan(x / 2) + C

    Explanation: Using trigonometric identity 1 + cos x = 2cos²(x / 2), and simplifying.

17. Question 17

    Q17. ∫(x + sin x) / (1 + cos x) dx

    • A) x tan(x / 2) + C
    • B) -x cot(x / 2) + C
    • C) x cot(x / 2) + C
    • D) -x tan(x / 2) + C

    Answer: x tan(x / 2) + C

    Explanation: Using substitution and trigonometric identities to simplify the integrand.

18. Question 18

    Q18. ∫x³ / √(1 - x²) dx on 0 to 1

    • A) 2 / 15
    • B) -2 / 15
    • C) 4 / 15
    • D) -4 / 15

    Answer: -2 / 15

    Explanation: Using substitution x = sin(θ), dx = cos(θ)dθ, and simplifying, then integrating within the given limits.

19. Question 19

    Q19. ∫1 / (x² + 2x + 5) dx

    • A) (1/2)tan^-1((x + 1) / 2) + C
    • B) (1/√2)tan^-1((x + 1) / √2) + C
    • C) (1/2)tan^-1((x + 1) / √2) + C
    • D) (1/√2)tan^-1((x + 1) / 2) + C

    Answer: (1/√2)tan^-1((x + 1) / √2) + C

    Explanation: Completing the square in the denominator, x² + 2x + 5 = (x + 1)² + 2², and using substitution u = (x + 1) / √2.

20. Question 20

    Q20. ∫(sin³x) / (cosx) dx = ?

    • A) (1/2)sin²x - ln|cosx| + C
    • B) (1/3)sin³x + C
    • C) -cosx + (1/3)cos³x + C
    • D) -(1/2)cos²x + ln|sinx| + C

    Answer: (1/2)sin²x - ln|cosx| + C

    Explanation: Using substitution u = cosx, du/dx = -sinx, and sin²x = 1 - cos²x, hence ∫(sin³x) / (cosx) dx = -∫(1 - u²)/u du.