Evaluate the following integrals:

let I $=\int \frac{1}{\mathrm{e}^{\mathrm{x}}+\mathrm{e}^{-\mathrm{x}}} \mathrm{dx}$

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

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

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

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

$I=\tan ^{-1} t+c$

$\left[\right.$ since, $\left.\int \frac{1}{1+(\mathrm{x})^{2}} \mathrm{dx}=\tan ^{-1} \mathrm{x}+\mathrm{c}\right]$

$I=\tan ^{-1}\left(e^{x}\right)+c[$ using $(i)]$