Integrate the rational function: $\frac{x}{(x-1)^{2}(x+2)}$

Let $\frac{x}{(x-1)^{2}(x+2)} = \frac{A}{(x-1)} + \frac{B}{(x-1)^{2}} + \frac{C}{(x+2)}$.
Multiplying both sides by $(x-1)^{2}(x+2)$,we get:
$x = A(x-1)(x+2) + B(x+2) + C(x-1)^{2}$.
For $x=1$,$1 = B(1+2) \Rightarrow 3B = 1 \Rightarrow B = \frac{1}{3}$.
For $x=-2$,$-2 = C(-2-1)^{2} \Rightarrow 9C = -2 \Rightarrow C = -\frac{2}{9}$.
Comparing the coefficients of $x^{2}$ on both sides: $0 = A + C \Rightarrow A = -C = \frac{2}{9}$.
Thus,$\frac{x}{(x-1)^{2}(x+2)} = \frac{2}{9(x-1)} + \frac{1}{3(x-1)^{2}} - \frac{2}{9(x+2)}$.
Integrating both sides:
$\int \frac{x}{(x-1)^{2}(x+2)} dx = \frac{2}{9} \int \frac{1}{x-1} dx + \frac{1}{3} \int (x-1)^{-2} dx - \frac{2}{9} \int \frac{1}{x+2} dx$.
$= \frac{2}{9} \log |x-1| + \frac{1}{3} \left( \frac{(x-1)^{-1}}{-1} \right) - \frac{2}{9} \log |x+2| + K$.
$= \frac{2}{9} \log \left| \frac{x-1}{x+2} \right| - \frac{1}{3(x-1)} + K$,where $K$ is the constant of integration.