Evaluate the following integrals:

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

We can write above integral as:

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

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

Let $e^{x}=t$

Differentiating w.r.t $\mathrm{x}$ we get,

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

$\therefore$ integral (1) becomes,

$=\tan ^{-1}(\mathrm{t})+C\left(\because \int \frac{1}{x^{2}+1} d x=\tan ^{-1}(x)\right)$

Putting value of $t$ we get,

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

$\therefore \int \frac{1}{e^{x}+e^{-x}} d x=\tan ^{-1}\left(e^{x}\right)+C$