Answer
$\frac{2x^2-x+8}{(x^2+4)^2}=\frac{2}{x^2+4}+\frac{-x}{(x^2+4)^2}$
Work Step by Step
$\frac{2x^2-x+8}{(x^2+4)^2}=\frac{Ax+B}{x^2+4}+\frac{Cx+D}{(x^2+4)^2}$,
$(Ax+B)(x^2+4)+Cx+D$,
$Ax^3+4Ax+Bx^2+4B+Cx+D$,
$Ax^3+Bx^2+(4A+C)x+4B+D$,
$\begin{cases}
A=0\\
B=2\\
4A+C=-1, C=-1\\
8+D=8, D=0
\end{cases}$
Therefore,
$\frac{2x^2-x+8}{(x^2+4)^2}=\frac{2}{x^2+4}+\frac{-x}{(x^2+4)^2}$
