Answer
$\frac{\sqrt {16-x^{2}}}{4x}+C$
Work Step by Step
$\int \frac{4}{x^{2}\sqrt {16-x^{2}}}dx$
Use $x=4\sin\theta$
$dx=4\cos\theta d\theta$
Use $\sin\theta =\frac{x}{4}$ for triangle.
$=\int \frac{16\cos\theta}{16\sin^{2}\theta\sqrt{16-16\sin^{2}\theta}}d\theta$
$=\int\frac{\cos\theta}{\sin^{2}\theta(4cos\theta)}d\theta$
$=\frac{1}{4}\int \csc^{2}\theta d\theta$
$=\frac{1}{4}\cot\theta+C$
Using the triangle, $\cot\theta =\frac{\sqrt {16-x^{2}}}{x}$
$=\frac{\sqrt {16-x^{2}}}{4x}+C$
