Answer
$$\frac{-\sqrt{4-x^2}}{4x}+c$$
Work Step by Step
Given
$$\int\frac{dx}{x^2\sqrt{4-x^2}} $$
Let $x=2\sin \theta \ \Rightarrow \ dx=2\cos \theta d \theta$, then
\begin{align*}
\int\frac{dx}{x^2\sqrt{4-x^2}}&=\int\frac{2\cos \theta d \theta}{4\sin^2 \theta\sqrt{4-4\sin^2 \theta}}\\
&=\int\frac{2\cos \theta d \theta}{4\sin^2 \theta(2\cos \theta)}\\
&=\frac{1}{4}\int \csc^2\theta d\theta\\
&=\frac{-1}{4}\cot \theta +c\\
&=\frac{-\sqrt{4-x^2}}{4x}+c
\end{align*}
