Evaluate the following integrals:

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

$e^{x}=t-1$

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

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

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

Substitute $t$ and dt we get

$\Rightarrow \int e^{2 x} \cdot \frac{d t}{e^{x}}$

$\Rightarrow \int e^{x} \cdot d t$

$\Rightarrow \int(t-1) d t$

$\Rightarrow \int t \cdot d t-\int d t$

$\Rightarrow \frac{t^{2}}{2}-t+c$

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

$\Rightarrow \frac{\left(1+e^{x}\right)^{2}}{2}-\left(1+e^{x}\right)+c$