Answer
The partial fraction is,
$\frac{x+1}{\left( {{x}^{2}}+2 \right)}-\frac{2x}{{{\left( {{x}^{2}}+2 \right)}^{2}}}$.
Work Step by Step
$\frac{{{x}^{3}}+{{x}^{2}}+2}{{{\left( {{x}^{2}}+2 \right)}^{2}}}=\frac{Ax+B}{\left( {{x}^{2}}+2 \right)}+\frac{Cx+D}{{{\left( {{x}^{2}}+2 \right)}^{2}}}$
Now, take the L.C.M. on the right side:
$\frac{{{x}^{3}}+{{x}^{2}}+2}{{{\left( {{x}^{2}}+2 \right)}^{2}}}=\frac{\left( Ax+B \right)\left( {{x}^{2}}+2 \right)+\left( Cx+D \right)}{{{\left( {{x}^{2}}+2 \right)}^{2}}}$
By eliminating the denominator from both sides, we get.
$\begin{align}
& {{x}^{3}}+{{x}^{2}}+2=\left( Ax+B \right)\left( {{x}^{2}}+2 \right)+\left( Cx+D \right) \\
& =A{{x}^{3}}+2Ax+B{{x}^{2}}+2B+Cx+D \\
& =A{{x}^{3}}+B{{x}^{2}}+\left( 2A+C \right)x+2B+D
\end{align}$
Then, compare the coefficients of ${{x}^{3}},{{x}^{2}},x$ and the constant term:
$A=1$ …… (I)
$B=1$ …… (II)
$2A+C=0$ …… (III)
$2B+D=2$ …… (IV)
And put the value of A in equation (III):
$\begin{align}
& 2+C=0 \\
& C=-2
\end{align}$
Also, put the value of B in equation (IV):
$\begin{align}
& 2\left( 1 \right)+D=2 \\
& D=2-2=0
\end{align}$
Then,
$\frac{{{x}^{3}}+{{x}^{2}}+2}{{{\left( {{x}^{2}}+2 \right)}^{2}}}=\frac{x+1}{\left( {{x}^{2}}+2 \right)}+\frac{-2x}{{{\left( {{x}^{2}}+2 \right)}^{2}}}$.
Thus, the partial fraction of the given expression is $\frac{x+1}{\left( {{x}^{2}}+2 \right)}-\frac{2x}{{{\left( {{x}^{2}}+2 \right)}^{2}}}$.
