Evaluate the following integrals:

First of all, take $e^{x}$ common from the denominator, so we get

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

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

Assume $e^{-x}+1=t$

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

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

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

Substituting $t$ and dt we get

$\Rightarrow \int \frac{-\mathrm{dt}}{\mathrm{t}}$

$\Rightarrow \ln |\mathrm{t}|+\mathrm{c}$

But $\mathrm{t}=\left(\mathrm{e}^{-\mathrm{x}}+1\right)$

$\Rightarrow \ln \left|e^{-x}+1\right|+c$