Answer
The partial fraction is, $\frac{2}{\left( x+1 \right)}+\frac{3x-1}{\left( {{x}^{2}}+2x+2 \right)}$.
$\frac{5{{x}^{2}}+6x+3}{\left( x+1 \right)\left( {{x}^{2}}+2x+2 \right)}=\frac{A}{\left( x+1 \right)}+\frac{Bx+C}{\left( {{x}^{2}}+2x+2 \right)}$
Work Step by Step
Now, take the L.C.M on the right side:
$\frac{5{{x}^{2}}+6x+3}{\left( x+1 \right)\left( {{x}^{2}}+2x+2 \right)}=\frac{A\left( {{x}^{2}}+2x+2 \right)+\left( Bx+C \right)\left( x+1 \right)}{\left( x+1 \right)\left( {{x}^{2}}+2x+2 \right)}$.
By eliminating the denominators from both sides, we get:
$\begin{align}
& 5{{x}^{2}}+6x+3=A\left( {{x}^{2}}+2x+2 \right)+\left( Bx+C \right)\left( x+1 \right) \\
& =A{{x}^{2}}+2Ax+2A+B{{x}^{2}}+Bx+Cx+C \\
& =\left( A+B \right){{x}^{2}}+\left( 2A+C+B \right)x+2A+C
\end{align}$
Then, compare the coefficient of ${{x}^{2}},\ x$ and constant term:
$A+B=5$ …… (I)
$2A+C+B=6$ …… (II)
$2A+C=3$ …… (III)
And put the value of equation (III) in equation (II):
$\begin{align}
& 3+B=6 \\
& B=3
\end{align}$
Also, put the value of B into equation (I):
$\begin{align}
& A+3=5 \\
& A=2
\end{align}$
Also, substitute the value of A in equation (III):
$\begin{align}
& 4+C=3 \\
& C=-1
\end{align}$
Then,
$\frac{5{{x}^{2}}+6x+3}{\left( x+1 \right)\left( {{x}^{2}}+2x+2 \right)}=\frac{2}{\left( x+1 \right)}+\frac{3x-1}{\left( {{x}^{2}}+2x+2 \right)}$.
Thus, the partial fraction of the provided expression is $\frac{2}{\left( x+1 \right)}+\frac{3x-1}{\left( {{x}^{2}}+2x+2 \right)}$.
