Answer
$-\ln (e^{-x}+1)+c$
$=x-\ln(1+e^x)+c$
Work Step by Step
Solve $\int \dfrac{dx}{1+e^x} $
This can be written as:$\int \dfrac{e^{-x}dx}{e^{-x}+1} $
Let us $p=e^{-x}+1$ and $dp=-e^{-p}dx$
This implies $\int \dfrac{e^{-x}dx}{e^{-x}+1} =-\int\dfrac{1}{p}dp$
$=-\ln |p|+c$
$=-\ln (e^{-x}+1)+c$
$=\ln (\dfrac{e^x}{1+e^x})+c$
Need to use logarithmic property, such as: $\ln m-\ln n=\ln (\dfrac{m}{n})$
Thus, $=\ln (e^x)-\ln(1+e^x)+c$
Hence, $=x-\ln(1+e^x)+c$
