Answer
$ \frac{1}{4x}+\frac{1}{x^2}-\frac{x+4}{4(x^2+4)}$
Work Step by Step
Step 1. Assume the rational function can be decomposed into
$\frac{x+4}{x^2(x^2+4)}=\frac{A}{x}+\frac{B}{x^2}+\frac{Cx+D}{x^2+4}$
Step 2. Evaluate the right side; we have
$RHS=\frac{Ax(x^2+4)+B(x^2+4)+x^2(Cx+D)}{x^2(x^2+4)}==\frac{(A+C)x^3+(B+D)x^2+(4A)x+4B}{x^2(x^2+4)}$
Step 3. Comparing with the LHS, we can set up the following:
$\begin{cases} A+C=0\\ B+D=0 \\ 4A=1 \\4B=4 \end{cases}$
Step 4. We can find the solutions as $A=\frac{1}{4}, B=1, C=-\frac{1}{4}, D=-1$
Step 5. The decomposition results is:
$\frac{x+4}{x^2(x^2+4)}=\frac{1}{4x}+\frac{1}{x^2}-\frac{x+4}{4(x^2+4)}$
