Answer
The partial fraction decomposition of the rational expression is $\frac{3}{5\left( x-3 \right)}+\frac{2}{5\left( x+2 \right)}$
Work Step by Step
Factor the denominator by the method of grouping as given below:
$\frac{x}{\left( x-3 \right)\left( x+2 \right)}=\frac{A}{x-3}+\frac{B}{x+2}$
Multiply both sides of the equation by $\left( x-3 \right)\left( x+2 \right)$ as given below:
$\begin{align}
& \left( x-3 \right)\left( x+2 \right)\cdot \frac{x}{\left( x-3 \right)\left( x+2 \right)}=\left( x-3 \right)\left( x+2 \right)\left( \frac{A}{x-3}+\frac{B}{x+2} \right) \\
& \left( x-3 \right)\left( x+2 \right)\cdot \frac{x}{\left( x-3 \right)\left( x+2 \right)}=\left( x-3 \right)\left( x+2 \right)\left( \frac{A}{x-3} \right)+\left( x-3 \right)\left( x+2 \right)\left( \frac{B}{x+2} \right)
\end{align}$
Divide the common factors:
$\begin{align}
& \left( x-3 \right)\left( x+2 \right)\cdot \frac{x}{\left( x-3 \right)\left( x+2 \right)}=\left( x-3 \right)\left( x+2 \right)\left( \frac{A}{x-3} \right)+\left( x-3 \right)\left( x+2 \right)\left( \frac{B}{x+2} \right) \\
& x=\left( x+2 \right)A+\left( x-3 \right)B \\
& x=Ax+2A+Bx-3B \\
& x=\left( A+B \right)x+\left( 2A-3B \right)
\end{align}$
Equate the coefficients of like terms of the equation to write a system of equations as given below:
$ A+B=1$ (I)
$2A-3B=0$ (II)
Multiply equation (I) by 3 and add with equation (II) to get the value of A:
$\begin{align}
& +\text{ }\underline{\begin{align}
& 3A+3B=3 \\
& 2A-3B=0 \\
\end{align}} \\
& \text{ }5A=3 \\
\end{align}$
Hence, the value of A is $\frac{3}{5}$.
Now, put the value of A in equation (I) so that the value of B can be computed as given below:
$\begin{align}
& A+B=1 \\
& \frac{3}{5}+B=1 \\
& B=1-\frac{3}{5} \\
& B=\frac{2}{5}
\end{align}$
Now, put the value of $ A $ and $ B $ in the initial equation and write the partial fraction decomposition as given below:
$\begin{align}
& \frac{x}{\left( x-3 \right)\left( x+2 \right)}=\frac{A}{x-3}+\frac{B}{x+2} \\
& =\frac{\frac{3}{5}}{x-3}+\frac{\frac{2}{5}}{x+2} \\
& =\frac{3}{5\left( x-3 \right)}+\frac{2}{5\left( x+2 \right)}
\end{align}$
Thus, the partial fraction decomposition of the rational expression is $\frac{3}{5\left( x-3 \right)}+\frac{2}{5\left( x+2 \right)}$.
