Answer
$\dfrac{x^3}{(x^2+4)^2}=\dfrac{x}{x^2+4}-\dfrac{-4x}{(x^2+4)^2}$
Work Step by Step
We are given the fraction:
$\dfrac{x^3}{(x^2+4)^2}$
Decompose the fraction:
$\dfrac{x^3}{(x^2+4)^2}=\dfrac{Ax+B}{x^2+4}+\dfrac{Cx+D}{(x^2+4)^2}$
Bring the fractions at the same denominator:
$\dfrac{x^3}{(x^2+4)^2}=\dfrac{(Ax+B)(x^2+4)}{(x^2+4)^2}+\dfrac{Cx+D}{(x^2+4)^2}$
$x^3=(Ax+B)(x^2+4)+(Cx+D)$
$x^3=Ax^3+4Ax+Bx^2+4B+Cx+D$
$x^3=Ax^3+Bx^2+(4A+C)x+D$
Identify the coefficients and build the system:
$\begin{cases}
A=1\\
B=0\\
4A+C=0\\
D=0
\end{cases}$
Determine $C$:
$4(1)+C=0$
$4+C=0$
$C=-4$
Write the decomposed fraction:
$\dfrac{x^3}{(x^2+4)^2}=\dfrac{x+0}{x^2+4}+\dfrac{-4x+0}{(x^2+4)^2}$
$\dfrac{x^3}{(x^2+4)^2}=\dfrac{x}{x^2+4}-\dfrac{-4x}{(x^2+4)^2}$
