Question
$\int \frac{d x}{e^{x}+e^{-x}}$ is equal to

Answer

Given Integral is: $\int \frac{d x}{e^{x}+e^{-x}}$
Let $I=\int \frac{d x}{e^{x}+e^{-x}}$
$= \int \frac{d x}{e^{-x}\left(e^{2 x}+1\right)}$
$= \int \frac{e^{x} d x}{\left(e^{2 x}+1\right)}$
Put $ e^{x }= t $
$\Rightarrow e^x dx = dt$
$\Rightarrow \int \frac{\mathrm{e}^{\mathrm{x}} \mathrm{d} \mathrm{x}}{\left(\mathrm{e}^{2 \mathrm{x}}+1\right)}=\int \frac{\mathrm{dt}}{\left(\mathrm{t}^{2}+1\right)}$
$= \tan^{-1} t + C$
$= \tan^{-1} (e^x) + C$

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