Evaluate the following integrals:

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

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

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

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

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

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

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

$\Rightarrow \frac{2 t^{3}}{3}+c$

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

$\Rightarrow \frac{2(1+\sqrt{x})^{3}}{3}+c$