How to find the integral of $\dfrac{1-x}{(1+x)^2}$? Can you help me with the $$ \int\frac{1-x}{(1+x)^2}dx$$ I was trying to integrate by parts but it doesn't work...

Substitution $u=1+x$ leads to: $$\int\frac{2-u}{u^{2}}du=2\int u^{-2}du-\int u^{-1}du=-2u^{-1}-\ln u+C=-2\left(1+x\right)^{-1}-\ln\left(1+x\right)+C$$
