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

$\tan ^{-1}\left(e^x\right)+c$, where c is a constant

The correct answer is Option (1) → $\tan ^{-1}\left(e^x\right)+c$, where c is a constant

$\int \frac{d x}{e^x+e^{-x}}=\int\frac{e^x}{1+(e^x)^2}dx$

let $y=e^x$

$dy=e^xdx$

$⇒\int\frac{dy}{1+x^2}=\tan^{-1}(y)+c$

$=\tan ^{-1}\left(e^x\right)+c$