Answer
$$\frac{2}{\sqrt{a}}\tan^{-1}\frac{\sqrt{x}}{\sqrt{a}}+C,\ \ \ \ \ \ a\neq0$$
$$\frac{-2}{\sqrt{x}}+C,\ \ \ \ \ \ a=0$$
Work Step by Step
Given $$\int \frac{d x}{x^{3 / 2}+a x^{1 / 2}}$$
Let $$ x=u^2 \ \ \ \ \ \ dx=2udu$$
Then for $a\neq 0$
\begin{align*}
\int \frac{d x}{x^{3 / 2}+a x^{1 / 2}}&=\int \frac{2ud u}{u^{3 }+a u}\\
&= \int \frac{2 d u}{u^{2 }+a }\\
&= \frac{2}{\sqrt{a}}\tan^{-1}\frac{u}{\sqrt{a}}+C\\
&= \frac{2}{\sqrt{a}}\tan^{-1}\frac{\sqrt{x}}{\sqrt{a}}+C
\end{align*}
For $a=0$
\begin{align*}
\int \frac{d x}{x^{3 / 2}}&=\frac{-2}{\sqrt{x}}+C
\end{align*}
