Answer
The partial fraction is, $\frac{3}{x}+\frac{2}{\left( x-1 \right)}-\frac{1}{\left( x+3 \right)}$
Work Step by Step
$\frac{4{{x}^{2}}+13x-9}{x\left( x-1 \right)\left( x+3 \right)}=\frac{A}{x}+\frac{B}{\left( x-1 \right)}+\frac{C}{\left( x+3 \right)}$
Now, multiply $x\left( x-3 \right)\left( 2x+1 \right)$ on both sides:
$\begin{align}
& x\left( x-1 \right)\left( x+3 \right)\frac{4{{x}^{2}}+13x-9}{x\left( x-1 \right)\left( x+3 \right)}=x\left( x-1 \right)\left( x+3 \right)\frac{A}{x}+x\left( x-1 \right)\left( x+3 \right)\frac{B}{\left( x-1 \right)}+x\left( x-1 \right)\left( x+3 \right)\frac{C}{\left( x+3 \right)} \\
& 4{{x}^{2}}+13x-9=\left( x-1 \right)\left( x+3 \right)A+x\left( x+3 \right)B+x\left( x-1 \right)C \\
& =\left( {{x}^{2}}+3x-x-3 \right)A+B{{x}^{2}}+3Bx+C{{x}^{2}}-Cx \\
& =A{{x}^{2}}+2Ax-3A+B{{x}^{2}}+3Bx+C{{x}^{2}}-Cx
\end{align}$ further simplify then,
$4{{x}^{2}}+13x-9={{x}^{2}}\left( A+B+C \right)+x\left( 2A+3B-C \right)-3A$ …… (1)
Then, compare the coefficient of equation (1):
$A+B+C=4$ …… (2)
$2A+3B-C=13$ …… (3)
$-3A=-9$ …… (4)
$A=3$ …… (5)
Now, put the value of A in equation (2):
$3+B+C=4$
$B+C=1$ …… (6)
Also, put the value of A in equation (3):
$2\left( 3 \right)+3B-C=13$
$3B-C=7$ …… (7)
Then, add equation (6) and equation (7), then
$B=2$
And put the value of B in equation (7):
$\begin{align}
& 3\left( 2 \right)-C=7 \\
& C=-1
\end{align}$
Therefore,
$\frac{4{{x}^{2}}+13x-9}{x\left( x-1 \right)\left( x+3 \right)}=\frac{3}{x}+\frac{2}{\left( x-1 \right)}-\frac{1}{\left( x+3 \right)}$
Thus, the partial fraction of the provided expression is $\frac{3}{x}+\frac{2}{\left( x-1 \right)}-\frac{1}{\left( x+3 \right)}$.
