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:
$\dfrac{x^3}{(x^2+4)^2}$
As the denominator is already factored, we can write the partial fraction decomposition:
$\dfrac{x^3}{(x^2+4)^2}=\dfrac{Ax+B}{x^2+4}+\dfrac{Cx+D}{(x^2+4)^2}$
Multiply all terms by the least common denominator $(x^2+4)^2$:
$(x^2+4)\cdot\dfrac{x^3}{(x^2+4)^2}=(x^2+4)\cdot\dfrac{Ax+B}{x^2+4}+(x^2+4)\cdot\dfrac{Cx+D}{(x^2+4)^2}$
$x^3=(Ax+B)(x^2+4)+(Cx+D)$
$x^3=Ax^3+4Ax+Bx^2+4B+Cx+D$
$x^3=Ax^3+Bx^2+(4A+C)x+(4B+D)$
Identify the coefficients and build the system of equations:
$\begin{cases}
A=1\\
B=0\\
4A+C=0\\
4B+D=0
\end{cases}$
Solve the system:
$A=1$
$B=0$
$4A+C=0$
$4(1)+C=0$
$C=-4$
$4B+D=0$
$4(0)+D=0$
$D=0$
The partial fraction decomposition is:
$\dfrac{x^3}{(x^2+4)^2}=\dfrac{x}{x^2+4}-\dfrac{4x}{(x^2+4)^2}$
