Answer
$\int \frac{dx}{x^{2}\sqrt{4-x^{2}}}=-\frac{\sqrt{4-x^{2}}}{4x}+C$
Work Step by Step
$$Let\ x=2\sin\,u,dx=2\cos\,u\,du,cot\,u=\frac{\sqrt{4-x^{2}}}{x}$$
$$\int \frac{dx}{x^{2}\sqrt{4-x^{2}}}=\int \frac{2\cos\,u\,du}{(2\sin\,u)^{2}\sqrt{4-(2\sin\,u)^{2}}}$$
$$=\int \frac{2\cos\,u}{4\sin^{2}u\,2\cos\,u}\,du=\frac{1}{4}\int \csc^{2}u\,du$$
$$=-\frac{1}{4}\cot\,u+C=-\frac{\sqrt{4-x^{2}}}{4x}+C$$
