Question:
Evaluate the following integrals:
$\int \frac{e^{x}}{1+e^{2 x}} d x$
Solution:
: 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)]