Answer
The partial fraction is, $\frac{2}{\left( x-3 \right)}+\frac{2x+5}{{{x}^{2}}+3x+3}$
Work Step by Step
Let us consider the equation:
$\frac{4{{x}^{2}}+5x-9}{{{x}^{3}}-6x-9x}$
$\frac{4{{x}^{2}}+5x-9}{\left( x-3 \right)\left( {{x}^{2}}+3x+3 \right)}$
Firstly, set up the partial fraction decomposition with the unknown constant:
$\frac{4{{x}^{2}}+5x-9}{\left( x-3 \right)\left( {{x}^{2}}+3x+3 \right)}=\frac{A}{\left( x-3 \right)}+\frac{Bx+C}{{{x}^{2}}+3x+3}$
And multiply both sides by: $\left( x-3 \right)\left( {{x}^{2}}+3x+3 \right)$:
$4{{x}^{2}}+5x+9=A\left( {{x}^{2}}+3x+3 \right)+Bx\left( x-3 \right)+C\left( x-3 \right)$ …… (I)
Then for the value of A, put $x=3$ in equation (I):
$\begin{align}
& 4\left( {{3}^{2}} \right)+5\left( 3 \right)-9=A\left( 9+9+3 \right) \\
& 36+15-9=21A \\
& 42=21A \\
& A=2s
\end{align}$
Put $x=0$ in equation (I):
$\begin{align}
& -9=3A-3C \\
& -9=3\left( 2 \right)-3C \\
& -9=6-3C \\
& 3C=15
\end{align}$
$C=5$
Then, put $x=1$ in equation (I):
$\begin{align}
& 4+5-9=7A-B-2C \\
& 0=14-2B-10 \\
& 2B=4 \\
& B=2
\end{align}$
Now, put the value of $A=2$ $B=2\text{ and }C=5$ in equation (I).
$\frac{4{{x}^{2}}+5x-9}{\left( x-3 \right)\left( {{x}^{2}}+3x+3 \right)}=\frac{2}{\left( x-3 \right)}+\frac{2x+5}{{{x}^{2}}+3x+3}$
Thus, the partial fraction is $\frac{2}{\left( x-3 \right)}+\frac{2x+5}{{{x}^{2}}+3x+3}$
