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

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Solution

$I = \int \frac{d x}{e^{x} + e^{- x}}$

$I = \int \frac{e^{x}}{e^{2 x} + 1} d x$

Let $e^{x} = t$
$e^{x} d x = d t$

$I = \int \frac{d t}{1 + t^{2}}$

$I = {tan}^{- 1} t + C$
$I = {tan}^{- 1} (e^{x}) + C$

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