Answer
The partial fraction decomposition of the rational expression is $\frac{x}{\left( x+1 \right)\left( {{x}^{2}}+9 \right)}=\frac{-1}{10\left( x+1 \right)}+\frac{x+9}{10\left( {{x}^{2}}+9 \right)}$.
Work Step by Step
It is required to compute the partial fraction decomposition. Factor the denominator by the method of grouping as shown below:
$\begin{align}
& \frac{x}{\left( x+1 \right)\left( {{x}^{2}}+9 \right)}=\frac{A}{x+1}+\frac{Bx+C}{{{x}^{2}}+9} \\
& =\frac{A\left( {{x}^{2}}+9 \right)+\left( Bx+C \right)\left( x+1 \right)}{\left( x+1 \right)\left( {{x}^{2}}+9 \right)}
\end{align}$
$ x=A\left( {{x}^{2}}+9 \right)+\left( Bx+C \right)\left( x+1 \right)$ (I)
Substitute the value of $ x=-1$ in equation (I), to obtain the value of A:
$\begin{align}
& -1=A\left( 10 \right) \\
& A=-\frac{1}{10}
\end{align}$
Again, putting the value of $ x=0$ in equation (I), we get,
$\begin{align}
& 0=A\left( 9 \right)+C\left( 1 \right) \\
& C=-9A \\
& C=-9.\left( -\frac{1}{10} \right) \\
& C=\frac{9}{10} \\
\end{align}$
Again, substitute the value of $ x=1$ in equation (I) as follows:
$\begin{align}
& 1=A\left( 10 \right)+\left( B+C \right)\left( 2 \right) \\
& 1=-\frac{1}{10}\left( 10 \right)+\left( B+\frac{9}{10} \right)2 \\
& 2=\left( B+\frac{9}{10} \right)2 \\
& 1=B+\frac{9}{10}
\end{align}$
$\begin{align}
& B=1-\frac{9}{10} \\
& B=\frac{1}{10}
\end{align}$
Therefore, the partial fraction decomposition of the rational expression is
$\frac{x}{\left( x+1 \right)\left( {{x}^{2}}+9 \right)}=\frac{-1}{10\left( x+1 \right)}+\frac{x+9}{10\left( {{x}^{2}}+9 \right)}$.
