Evaluate the following integrals:

Assume $\sqrt{x}=\mathrm{t}$

$d(\sqrt{x})=d t$

$\Rightarrow \frac{1}{2 \sqrt{x}} \mathrm{dx}=\mathrm{dt}$

$\Rightarrow \frac{1}{\sqrt{x}} \mathrm{dx}=2 \mathrm{dt}$

Substituting $t$ and $d t$

$\Rightarrow 2 \int \sec ^{2} t d t$

$=2 \tan t+c$

But $\sqrt{x}=t$

$\Rightarrow 2 \tan (\sqrt{x})+c$