Question

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

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \frac{1}{e^x+e^{-x}}dx \\ =\int \frac{dx}{e^x+{\displaystyle \frac{1}{e^x}}} \\ =\int \frac{e^{x}dx}{e^{2x}+1} \\ =\int \frac{e^{x}dx}{{\left(e^x\right)}^2+1} \\ \mathrm{Putting}{e}^x=t \\ \Rightarrow e^{x}dx=dt \\ \therefore I=\int \frac{dt}{t^2+1} \\ ={\tan }^{-1}t+C\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\left(\because \int \frac{dt}{a^{\mathit{2}}\mathit{+}{x}^{\mathit{2}}}=\frac{1}{a}{\tan }^{\mathit{-}1}\frac{x}{a}+C\right) \\ ={\tan }^{-1}{e}^x+C\mathit{}\left(\because t=e^x\right) \end{gathered}$$