Answer
$$\int \frac{x}{(x^2-4)^3}dx=-\frac{1}{4(x^2-4)^2}+C$$
Work Step by Step
$$A=\int \frac{x}{(x^2-4)^3}dx$$
We set $u=x^2-4$
Then $$du=2xdx$$ $$xdx=\frac{1}{2}du$$
Therefore, $$A=\frac{1}{2}\int\frac{1}{u^3}du=\frac{1}{2}\int u^{-3}du$$ $$A=\frac{1}{2}\times\frac{u^{-2}}{-2}+C$$ $$A=-\frac{u^{-2}}{4}+C=-\frac{1}{4u^2}+C$$ $$A=-\frac{1}{4(x^2-4)^2}+C$$
