Question

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

Answer 1

Let $\frac{x}{(x-1{)}^2(x+2)}=\frac{A}{(x-1)}+\frac{B}{(x-1{)}^2}+\frac{C}{(x+2)}$

$\Rightarrow x=A(x-1)(x+2)+B(x+2)+C(x-1{)}^2$

Substituting $x=1$, we obtain $B=\frac{1}{3}$

Equating the coefficients of $x^2$ and constant term, we obtain
$A+C=0$

$2A+2B+C=0$

On solving, we obtain

$A=\frac{2}{9}$ and $C=\frac{-2}{9}$

$\therefore \frac{x}{(x-1{)}^2(x+2)}=\frac{2}{9(x-1)}+\frac{1}{3(x-1{)}^2}-\frac{2}{9(x+2)}$

$\Rightarrow \int \frac{x}{(x-1{)}^2(x+2)}dx=\frac{2}{9}\int \frac{1}{(x-1)}dx+\frac{1}{3}\int \frac{1}{(x-1{)}^2}dx-\frac{2}{9}\int \frac{1}{(x+2)}dx$

$=\frac{2}{9}\log |x-1|+\frac{1}{3}\left(\frac{-1}{x-1}\right)-\frac{2}{9}\log |x+2|+C$

$=\frac{2}{9}\log \left|\frac{x-1}{x+2}\right|-\frac{1}{3(x-1)}+C$