Evaluate the following integrals:

Assume $e^{2 x}-2=t$

$d\left(e^{2 x}-2\right)=d t$

$\Rightarrow 2 e^{2 x} d x=d t$

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

Put $\mathrm{t}$ and $\mathrm{dt}$ in the given equation we get

$\Rightarrow \frac{1}{2} \int \frac{\mathrm{dt}}{\mathrm{t}}$

$=\frac{1}{2} \ln |\mathrm{t}|+\mathrm{c}$

But $t=e^{2 x}-2$

$=\frac{1}{2} \ln \left|e^{2 x}-2\right|+c$