Answer
$\displaystyle \frac{4}{3}(\sqrt{x}+1)^{3/2}-4\sqrt{\sqrt{x}+1}+C$.
Work Step by Step
$I=\displaystyle \int\frac{1}{\sqrt{\sqrt{x}+1}}dx=$ sub $\left[\begin{array}{ll}
u=\sqrt{x}+1 & x=(u-1)^{2}\\
& dx=2(u-1)du
\end{array}\right]$
=$\displaystyle \int\frac{2(u-1)du}{\sqrt{u}}$
=$\displaystyle \int\frac{2udu}{\sqrt{u}}-\int\frac{2du}{\sqrt{u}}$
$=2\displaystyle \int u^{1/2}du-2\int u^{-1/2}du$
... both are table integrals type 1.
$=2(\displaystyle \frac{2}{3}u^{3/2})-2(2u^{1/2})+C$
... bring back x
$=\displaystyle \frac{4}{3}(\sqrt{x}+1)^{3/2}-4\sqrt{\sqrt{x}+1}+C$
