Answer
$$4\left(\frac{1}{3}(\sqrt{x}+1)^{3/2}- \sqrt{x}-1 \right)+C$$
Work Step by Step
Given
$$ \int \frac{d x}{\sqrt{\sqrt{x}+1}}$$
Let
$$u^2=\sqrt{x}+1 \ \ \ \ \ 2udu =\frac{1}{2\sqrt{x}}dx $$
Then
$$ dx=4\sqrt{x}udu\ \ \to \ dx= 4(u^2-1)udu$$
Hence
\begin{aligned}
\int \frac{d x}{\sqrt{\sqrt{x}+1}}&=\int \frac{4(u^2-1)udu}{\sqrt{u^2}}\\
&= \int 4(u^2-1)du\\
&=4\left(\frac{1}{3}u^3- u \right)+C\\
&= 4\left(\frac{1}{3}(\sqrt{x}+1)^{3/2}- \sqrt{x}-1 \right)+C
\end{aligned}
