More practice questions

  1. Q5. Find ∫2x/(x2 + 3x + 2) dx (NCERT Exercise 7.5, Q5, page 258).

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    x2 + 3x + 2 = (x + 1)(x + 2), and 2x = A(x + 2) + B(x + 1) gives A = −2 (x = −1) and B = 4 (x = −2). The integral is 4 log |x + 2| − 2 log |x + 1| + C.

  2. Q6. Find ∫(1 − x2)/(x(1 − 2x)) dx (NCERT Exercise 7.5, Q6, page 258).

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    Top and bottom both have degree 2, so divide: (1 − x2)/(x − 2x2) = ½ + (1 − x/2)/(x(1 − 2x)). The proper part is A/x + B/(1 − 2x) with A = 1 (x = 0) and B = 3/2 (x = ½), and ∫dx/(1 − 2x) = −½ log |1 − 2x|. The integral is x/2 + log |x| − (3/4) log |1 − 2x| + C.

  3. Q7. Find ∫5x/((x + 1)(x2 − 4)) dx (NCERT Exercise 7.5, Q11, page 258).

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    x2 − 4 = (x − 2)(x + 2), so there are three linear factors. Putting x = −1, 2 and −2 gives the coefficients 5/3, 5/6 and −5/2. The integral is (5/3) log |x + 1| + (5/6) log |x − 2| − (5/2) log |x + 2| + C.

  4. Q8. Find ∫dx/(x4 − 1) (NCERT Exercise 7.5, Q15, page 258).

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    1/(x4 − 1) = ½[1/(x2 − 1) − 1/(x2 + 1)], since the difference of the two fractions is 2/(x4 − 1). So the integral is ¼ log |(x − 1)/(x + 1)| − ½ tan x + C.

  5. Q9. Find ∫dx/(x(xn + 1)) by multiplying the top and bottom by xn − 1 and putting t = xn (NCERT Exercise 7.5, Q16, page 258).

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    For n ≠ 0, on an interval where the powers are real and the original denominator is nonzero, dt = nxn − 1 dx, so the integral is (1/n)∫dt/(t(t + 1)) = (1/n)∫[1/t − 1/(t + 1)] dt = (1/n) log |xn/(xn + 1)| + C. If n = 0, integrate 1/(2x) directly to obtain ½ log |x| + C.

  6. Q10. Find the integral of cos x divided by (1 − sin x)(2 − sin x) (NCERT Exercise 7.5, Q17, page 258).

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    Put t = sin x: ∫dt/((1 − t)(2 − t)) = ∫[1/(1 − t) − 1/(2 − t)] dt = −log |1 − t| + log |2 − t|. The answer is log |(2 − sin x)/(1 − sin x)| + C.

  7. Q11. Find ∫2x/((x2 + 1)(x2 + 3)) dx (NCERT Exercise 7.5, Q19, page 259).

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    Put t = x2, so dt = 2x dx: ∫dt/((t + 1)(t + 3)) = ½∫[1/(t + 1) − 1/(t + 3)] dt. The answer is ½ log ((x2 + 1)/(x2 + 3)) + C.

  8. Q12. Find ∫dx/(ex − 1) (NCERT Exercise 7.5, Q21, page 259).

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    Put t = ex, so dx = dt/t: ∫dt/(t(t − 1)) = ∫[1/(t − 1) − 1/t] dt = log |(t − 1)/t|. The answer is log |(ex − 1)/ex| + C.