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...

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 $2(1 + x)^{-1}$
