Evaluate the following integrals:

: let $\mathrm{I}=\int \frac{\mathrm{e}^{\mathrm{x}}}{1+\mathrm{e}^{2 \mathrm{x}}} \mathrm{dx}$

Let $e^{x}=t \ldots \ldots$ (i)

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

SO,

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

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

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

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