#### Answer

$\dfrac{x^3}{(x^2+4)^2}=\dfrac{x}{x^2+4}-\dfrac{4x}{(x^2+4)^2}$

#### Work Step by Step

Here, we have $\dfrac{x^3}{(x^2+4)^2}=\dfrac{Ax+B}{x^2+4}+\dfrac{Cx+D}{(x^2+4)^2}$
$x^3=Ax^3+Bx^2+(4A+C)x+4B+D$
Equating the coefficients, we get
$A=1$ and $B=0$
and $4A+C=0$ and $4B+D=0$
This gives: $A=1; B=0; C= -4; D=0$
Hence, $\dfrac{x^3}{(x^2+4)^2}=\dfrac{x}{x^2+4}-\dfrac{4x}{(x^2+4)^2}$