Answer
$$
\int \frac{1}{\sqrt{x+x^{3 / 2}}} d x =4 \sqrt{1+\sqrt{x}}+C
$$
Work Step by Step
$$
\int \frac{1}{\sqrt{x+x^{3 / 2}}} d x =\int \frac{d x}{\sqrt{x(1+\sqrt{x})}}=\int \frac{d x}{\sqrt{x} \sqrt{1+\sqrt{x}}}
$$
Let $ u = 1+\sqrt{x}$. Then $ du = \frac{x}{2\sqrt{x}}$ , so
$$
\begin{aligned}
\int \frac{1}{\sqrt{x+x^{3 / 2}}} d x &=\int \frac{2 d u}{\sqrt{u}} =\int 2 u^{-1 / 2} d u\\
&=4 \sqrt{u}+C=4 \sqrt{1+\sqrt{x}}+C
\end{aligned}
$$
where $C$ is constant