Answer
The partial fraction decomposition of the rational expression is $\frac{x}{\left( {{x}^{2}}+4 \right)}-\frac{4x}{{{\left( {{x}^{2}}+4 \right)}^{2}}}$
Work Step by Step
We know that the expression ${{x}^{2}}+4$ occurs multiple times in the denominator of the rational expression; therefore, for each power of ${{x}^{2}}+4$ assign an undefined constant factor in the denominator by the method of grouping as shown below:
$\frac{{{x}^{3}}}{{{\left( {{x}^{2}}+4 \right)}^{2}}}=\frac{Ax+B}{\left( {{x}^{2}}+4 \right)}+\frac{Cx+D}{{{\left( {{x}^{2}}+4 \right)}^{2}}}$
Multiply both sides of the equation by ${{\left( {{x}^{2}}+4 \right)}^{2}}$:
$\begin{align}
& {{\left( {{x}^{2}}+4 \right)}^{2}}\cdot \frac{{{x}^{3}}}{{{\left( {{x}^{2}}+4 \right)}^{2}}}={{\left( {{x}^{2}}+4 \right)}^{2}}\left( \frac{Ax+B}{\left( {{x}^{2}}+4 \right)}+\frac{Cx+D}{{{\left( {{x}^{2}}+4 \right)}^{2}}} \right) \\
& {{\left( {{x}^{2}}+4 \right)}^{2}}\cdot \frac{{{x}^{3}}}{{{\left( {{x}^{2}}+4 \right)}^{2}}}={{\left( {{x}^{2}}+4 \right)}^{2}}\left( \frac{Ax+B}{\left( {{x}^{2}}+4 \right)} \right)+{{\left( {{x}^{2}}+4 \right)}^{2}}\left( \frac{Cx+D}{{{\left( {{x}^{2}}+4 \right)}^{2}}} \right)
\end{align}$
And divide out the common factors:
$\begin{align}
& {{\left( {{x}^{2}}+4 \right)}^{2}}\cdot \frac{{{x}^{3}}}{{{\left( {{x}^{2}}+4 \right)}^{2}}}=\left( {{x}^{2}}+4 \right)\left( {{x}^{2}}+4 \right)\left( \frac{Ax+B}{\left( {{x}^{2}}+4 \right)} \right)+{{\left( {{x}^{2}}+4 \right)}^{2}}\left( \frac{Cx+D}{{{\left( {{x}^{2}}+4 \right)}^{2}}} \right) \\
& {{x}^{3}}=\left( {{x}^{2}}+4 \right)\left( Ax+B \right)+\left( Cx+D \right) \\
& {{x}^{3}}=A{{x}^{3}}+4Ax+B{{x}^{2}}+4B+Cx+D \\
& {{x}^{3}}=A{{x}^{3}}+B{{x}^{2}}+\left( 4A+C \right)x+\left( 4B+D \right)
\end{align}$
Then, equate the coefficients of like terms of the equation to write a system of equations as shown below:
$ A=1$ (I)
$ B=0$ (II)
$4A+C=0$ (III)
$4B+D=0$ (IV)
Put $ A=1$ in equation (III) to obtain the value of C:
$\begin{align}
& 4A+C=0 \\
& 4\left( 1 \right)+C=0 \\
& C=-4
\end{align}$
Put $ B=0$ in equation (IV) to obtain the value of D:
$\begin{align}
& 4B+D=0 \\
& 4\left( 0 \right)+D=0 \\
& D=0
\end{align}$
Put the values of $ A,\,\,B,\,\,C,\,\,\text{and }D $ in the given equation, and define the partial fraction decomposition:
$\begin{align}
& \frac{{{x}^{3}}}{{{\left( {{x}^{2}}+4 \right)}^{2}}}=\frac{Ax+B}{\left( {{x}^{2}}+4 \right)}+\frac{Cx+D}{{{\left( {{x}^{2}}+4 \right)}^{2}}} \\
& =\frac{\left( 1 \right)x+0}{\left( {{x}^{2}}+4 \right)}+\frac{\left( -4 \right)x+0}{{{\left( {{x}^{2}}+4 \right)}^{2}}} \\
& =\frac{x}{\left( {{x}^{2}}+4 \right)}-\frac{4x}{{{\left( {{x}^{2}}+4 \right)}^{2}}}
\end{align}$
Hence, the partial fraction decomposition of the rational expression is
$\frac{x}{\left( {{x}^{2}}+4 \right)}-\frac{4x}{{{\left( {{x}^{2}}+4 \right)}^{2}}}$.
