Answer
$$\frac{-\sqrt{4-x^{2}}}{4 x}+C$$
Work Step by Step
Given $$\int \frac{d x}{x^{2} \sqrt{4-x^{2}}}$$
Let
$$x=2\sin u\ \ \ \ \ \ \ dx= 2\cos udu $$
Then
\begin{align*}
\int \frac{d x}{x^{2} \sqrt{4-x^{2}}}&=\int \frac{2\cos udu}{4\sin^{2}u \sqrt{4-4\sin^{2}u}}\\
&=\int \frac{2\cos udu}{4\sin^{2}u \sqrt{4cos^{2}u}}\\
&=\frac{1}{4}\int \csc^2 udu\\
&=\frac{-1}{4}\cot u+C\\
&=\frac{-\sqrt{4-x^{2}}}{4 x}+C
\end{align*}
