Evaluate the following integrals:

Assume $1+\mathrm{e}^{\mathrm{x}}=\mathrm{t}$

$\Rightarrow \mathrm{d}\left(1+\mathrm{e}^{\mathrm{x}}\right)=\mathrm{dt}$

$\Rightarrow \mathrm{e}^{\mathrm{x}} \mathrm{dx}=\mathrm{dt}$

$\therefore$ Substituting $t$ and $d t$ in given equation we get

$\Rightarrow \int \frac{1}{t^{2}} d t$

$\Rightarrow \int t^{-2} \cdot d t$

$\Rightarrow \frac{-1}{t}+c$

But $1+e^{x}=t$

$\Rightarrow \frac{-1}{1+e^{x}}+c$