Show that

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

$\Rightarrow x=A(x+2)+B(x+1)$

Equating the coefficients of x and constant term, we obtain

$A+B=1$

$2 A+B=0$

On solving, we obtain

$A=-1$ and $B=2$

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

$\Rightarrow \int \frac{x}{(x+1)(x+2)} d x=\int \frac{-1}{(x+1)}+\frac{2}{(x+2)} d x$

$=-\log |x+1|+2 \log |x+2|+C$

$=\log (x+2)^{2}-\log |x+1|+C$

$=\log \frac{(x+2)^{2}}{(x+1)}+C$