Answer
The partial fraction is, $\frac{2}{\left( x-2 \right)}+\frac{-2x+1}{\left( {{x}^{2}}+2x+2 \right)}$.
Work Step by Step
$\frac{9x+2}{\left( x-2 \right)\left( {{x}^{2}}+2x+2 \right)}=\frac{A}{\left( x-2 \right)}+\frac{Bx+C}{\left( {{x}^{2}}+2x+2 \right)}$.
Now, take the L.C.M on the right side:
$\frac{9x+2}{\left( x-2 \right)\left( {{x}^{2}}+2x+2 \right)}=\frac{A\left( {{x}^{2}}+2x+2 \right)+\left( Bx+C \right)\left( x-2 \right)}{\left( x-2 \right)\left( {{x}^{2}}+2x+2 \right)}$.
By eliminating the denominators from both sides, we get:
$\begin{align}
& 9x+2=A\left( {{x}^{2}}+2x+2 \right)+\left( Bx+C \right)\left( x-2 \right) \\
& =A{{x}^{2}}+2Ax+2A+B{{x}^{2}}-2Bx+Cx-2C \\
& =\left( A+B \right){{x}^{2}}+\left( 2A+C-2B \right)x+2A-2C
\end{align}$
Then, compare the coefficient of ${{x}^{2}},\ x$ and constant term:
$A+B=0$ …… (I)
$2A+C-2B=9$ …… (II)
$2A-2C=2$
$A-C=1$ …… (III)
Now, find the values of A and B from equation (III) and equation (I) respectively and put in equation (II):
$\begin{align}
& A=1+C \\
& B=-A \\
& B=-C-1 \\
\end{align}$
$\begin{align}
& 2\left( 1+C \right)+C-2\left( -C-1 \right)=9 \\
& 2+2C+C+2C+2=9 \\
& 5C=5 \\
& C=1
\end{align}$
Then, put the value of C in equation (III):
$\begin{align}
& A-1=1 \\
& A=2
\end{align}$
And put the value of A in equation (I):
$\begin{align}
& 2+B=0 \\
& B=-2
\end{align}$
Then,
$\frac{9x+2}{\left( x-2 \right)\left( {{x}^{2}}+2x+2 \right)}=\frac{2}{\left( x-2 \right)}+\frac{-2x+1}{\left( {{x}^{2}}+2x+2 \right)}$
Thus, the partial fraction of the provided expression is $\frac{2}{\left( x-2 \right)}+\frac{-2x+1}{\left( {{x}^{2}}+2x+2 \right)}$.
