\(\int\displaystyle\frac{x}{(x-1)^2(x+2)}\,dx\) is equal to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{2}{9}\log\left|\frac{x-1}{x+2}\right|-\frac{1}{3(x-1)}+C\)
For the repeated factor \((x-1)^2\), the partial fractions are
\( \displaystyle \frac{x}{(x-1)^2(x+2)} = \frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{D}{x+2} \)
\( x = A(x-1)(x+2)+B(x+2)+D(x-1)^2 \)
Putting x = 1:
\( \displaystyle 1 = 3B,\quad B = \frac{1}{3} \)
Putting x = −2:
\( \displaystyle -2 = 9D,\quad D = -\frac{2}{9} \)
Comparing the coefficients of \(x^2\):
\( \displaystyle 0 = A+D,\quad A = \frac{2}{9} \)
Integrating each part, with \(\int\frac{dx}{(x-1)^2} = -\frac{1}{x-1}\):
\( \displaystyle I = \frac{2}{9}\log|x-1|-\frac{1}{3(x-1)}-\frac{2}{9}\log|x+2|+C \)
\( \displaystyle = \frac{2}{9}\log\left|\frac{x-1}{x+2}\right|-\frac{1}{3(x-1)}+C \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{2}{9}\log\left|\frac{x-1}{x+2}\right|+\frac{1}{3(x-1)}+C\) — Incorrect. \(\int\frac{dx}{(x-1)^2} = -\frac{1}{x-1}\), so this term is negative.
Option (3) \(\displaystyle\frac{2}{9}\log\left|\frac{x+2}{x-1}\right|-\frac{1}{3(x-1)}+C\) — Incorrect. The fraction inside the log is upside down, since \(A = \frac{2}{9}\) goes with \(x-1\).
Option (4) \(\displaystyle\frac{1}{9}\log\left|\frac{x-1}{x+2}\right|-\frac{1}{3(x-1)}+C\) — Incorrect. The log coefficient is \(A = \frac{2}{9}\), not \(\frac{1}{9}\).