Let $\frac{x}{(x-1)^2(x+2)}$ $= \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+2}$ ⇒ x = A(x – 1)(x + 2) + B(x + 2) + C(x – 1)2 ...(i) Comparing coefficient of $x^2$ on both sides of (i), we get, 0 = A + C Putting x = –2 in (i), we get $-2 = C(-2-1)^2$ $\Rightarrow C = -\frac{2}{9}$ $\Rightarrow A = -C + \frac{2}{9}$Putting x = 1 in (i), we get 1 = B(1 + 2) ⇒ B $= \frac{1}{3}$ $= \frac{2}{9} \log \left| \frac{x-1}{x+2} \right| - \frac{1}{3(x-1)} + C$