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= t $\Rightarrow$ ex 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 (ex) + C

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