Answer
$\frac{ax^3+bx^2}{(x^2+1)^2}=\frac{ax+b}{x^2+1}+\frac{-ax-b}{(x^2+1)^2}$,
Work Step by Step
$\frac{ax^3+bx^2}{(x^2+1)^2}=\frac{Ax+B}{x^2+1}+\frac{Cx+D}{(x^2+1)^2}$,
$(Ax+B)(x^2+1)+Cx+D$,
$Ax^3+Ax+Bx^2+B+Cx+D$,
$Ax^3+Bx^2+(A+C)x+B+D=ax^3+bx^2$,
$\begin{cases}
A=a\\
B=b\\
A+C=0\\
B+D=0
\end{cases}$,
thus, $a+C=0, C=-a$, substituting back into the Equation, $b+D=0, D=-b$,
Therefore,
$\frac{ax^3+bx^2}{(x^2+1)^2}=\frac{ax+b}{x^2+1}+\frac{-ax-b}{(x^2+1)^2}$,
