Question

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

Answer 1

$$\begin{gathered} \int \frac{dx}{e^x+e^{-x}} \\ =\int \frac{dx}{e^x+{\displaystyle \frac{1}{e^x}}} \\ =\int \frac{e^{x}dx}{e^{2x}+1} \\ \mathrm{let}{e}^x=t \\ \Rightarrow e^{x}dx=dt \\ \mathrm{Now},\int \frac{e^{x}dx}{e^{2x}+1} \\ =\int \frac{dt}{1+t^2} \\ ={\tan }^{-1}\left(t\right)+c \\ ={\tan }^{-1}\left(e^x\right)+c \end{gathered}$$