Use simple and repeated linear factors

$\frac1{(1+x)(2-x)}=\frac13\bigl(\frac1{1+x}+\frac1{2-x}\bigr)$, so the integral is $\frac13[\ln|1+x|-\ln|2-x|]+C=\frac13\ln\left|\frac{1+x}{2-x}\right|+C$, for $x\ne-1,2$.

Write $x=(x+1)-1$ to get $x/(x+1)^2=1/(x+1)-1/(x+1)^2$. Integrating gives $\ln|x+1|+1/(x+1)+C$, for $x\ne-1$, matching the printed answer.