Answer
$\displaystyle \frac{x^{3}+x^{2}}{(x^{2}+4)^2}=\frac{Ax+B}{x^2+4}+\frac{Cx+D}{(x^{2}+4)}$
Work Step by Step
The factor $(x^2+4)$ is a repeated quadratic factor,
see p.548. Repeated Prime Quadratic Factor In the Denominator
Include one partial fraction with a $linear$ numerator
for each $power$ of a prime, repeated quadratic factor in the denominator
Setup only:
$\displaystyle \frac{x^{3}+x^{2}}{(x^{2}+4)^2}=\frac{Ax+B}{x^2+4}+\frac{Cx+D}{(x^{2}+4)^2}$
