Answer
$\frac{-1}{4}(x^2-4)^{-2}+C$
Work Step by Step
Let $u=x^2-4=>du=2x dx$
and $\frac{1}{2}du=xdx$
thus $\int \frac{x}{(x^2-4)^3}dx$
=$\int (x^2-4)^{-3}xdx$
=$\int u^{-3}\frac{1}{2}du=\frac{1}{2}\int u^{-3}du$
=$\frac{-1}{4}u^{-2}+C$
=$\frac{-1}{4}(x^2-4)^{-2}+C$
