Let $\frac{2 x}{x^{2}+3 x+2}=\frac{A}{(x+1)}+\frac{B}{(x+2)}$

$2 x=A(x+2)+B(x+1)$   ...(1)

Substituting x = −1 and −2 in equation (1), we obtain

$A=-2$ and $B=4$

$\therefore \frac{2 x}{(x+1)(x+2)}=\frac{-2}{(x+1)}+\frac{4}{(x+2)}$

$\Rightarrow \int \frac{2 x}{(x+1)(x+2)} d x=\int\left\{\frac{4}{(x+2)}-\frac{2}{(x+1)}\right\} d x$

$=4 \log |x+2|-2 \log |x+1|+\mathrm{C}$