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