Question

$\int \frac{1}{e^x+1}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \frac{1}{e^x+1}dx \\ =\int \frac{e^{-x}}{1+e^{-x}}dx \\ \mathrm{Putting}{e}^{-x}=t \\ \Rightarrow -e^{-x}=\frac{dt}{dx} \\ \Rightarrow e^{-x}dx=-dt \\ \therefore I=\int \frac{-1}{1+t}dt \\ =-\ln \left|1+t\right|+\mathrm{C} \\ =-\ln \left|1+e^{-x}\right|+\mathrm{C} \end{gathered}$$