Evaluate the following integrals:

Assume $\sqrt{x}+1=t$

$d(\sqrt{x}+1)=d t$

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

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

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

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

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

But $t=\sqrt{x}+1$

$=2 \ln |\sqrt{x}+1|+c$