Answer
$$\frac{4}{5}\left(\sqrt{1+\sqrt{x}}\right)^{5}- \frac{4}{3}\left(\sqrt{1+\sqrt{x}}\right)^{3}+C$$
Work Step by Step
Given $$\int \sqrt{1+\sqrt{x}} d x$$
Let
\begin{aligned}
u &=\sqrt{1+\sqrt{x}} \\
u^{2} &=1+\sqrt{x} \\
\sqrt{x} &=u^{2}-1
\end{aligned}
Then $ d x=4 u\left(u^{2}-1\right) d u$, and hence
\begin{aligned}
\int \sqrt{1+\sqrt{x}} d x &=4 \int u^{2}\left(u^{2}-1\right) d u \\
&=4 \int\left(u^{4}-u^{2}\right) d u\\
&= 4 \frac{u^{5}}{5}-4 \frac{u^{3}}{3}+C\\
&= \frac{4}{5}\left(\sqrt{1+\sqrt{x}}\right)^{5}- \frac{4}{3}\left(\sqrt{1+\sqrt{x}}\right)^{3}+C
\end{aligned}
