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CA Foundation · Quantitative Aptitude · Differential and Integral Calculus

Evaluate ∫ 1/((x − 1)(x + 2)) dx, where C is the constant of integration.

Splitting into partial fractions gives (1/3)[1/(x−1) − 1/(x+2)]. Integrating each term yields (1/3)log|x−1| − (1/3)log|x+2| + C, which equals (1/3)log|(x−1)/(x+2)| + C.

  1. A(1/3) log|(x − 1)/(x + 2)| + CCorrect
  2. B(1/3) log|(x + 2)/(x − 1)| + C
  3. Clog|(x − 1)(x + 2)| + C
  4. D(1/3) log|(x − 1)(x + 2)| + C

Explanation

Write 1/((x−1)(x+2)) = A/(x−1) + B/(x+2). Then 1 = A(x+2) + B(x−1); x = 1 gives A = 1/3, x = −2 gives B = −1/3. Integrating gives (1/3)log|x−1| − (1/3)log|x+2| + C = (1/3)log|(x−1)/(x+2)| + C. The inverted ratio option has the wrong sign for B.

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