2015/067) Integrate $\dfrac{1}{e^x(1+e^x)}$

1st we convert it into partial fraction
$\dfrac{1}{e^x} - \dfrac{1}{e^x +1}$
= $e^{-x} - \dfrac{e^{-x}}{1+ e^{-x}}$
integral of $e^{-x}$ is $-e^{-x}$

and putting $1+e^{-x} = t$ we get $-e^{-x} dx = dt$
so integral of $\dfrac{e^{-x}}{1+ e^{-x}}$ = $ln | 1 + e^{-x} |$
but as $1 + e^{-x} > 1$ and hence > 0 we get the integral of given expression

$- e^{- x} + ln ( 1+ e^{-x}) + C$