How to find the integral of $\dfrac{1-x}{(1+x)^2}$?

Question

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...

Answer 1

Write: $1 - x = 1 + x - 2x$, and split it into 2 parts:

Part 1: $\dfrac{(1 + x )}{( 1 + x)^2} = \dfrac1{(1 + x)}$ whose anti derivative is $\ln(1 + x)$

Part 2:

$-2x = -2(1 + x) + 2$, and $\dfrac{-2(1 + x)}{(1 + x)^2} = \dfrac{-2}{1 + x}$ with \anti derivative $-2ln(1 + x)$, and the other part $\dfrac{2}{(1 + x)^2}$ has anti derivative $2(1 + x)^{-1}$

Answer 2

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