Answer
The partial fraction is, $\frac{2}{\left( x+1 \right)}+\frac{4}{{{\left( x+1 \right)}^{2}}}-\frac{3}{{{\left( x+1 \right)}^{3}}}$
Work Step by Step
$\frac{2{{x}^{2}}+8x+3}{{{\left( x+1 \right)}^{3}}}=\frac{A}{\left( x+1 \right)}+\frac{B}{{{\left( x+1 \right)}^{2}}}+\frac{C}{{{\left( x+1 \right)}^{3}}}$
Now, multiply both sides by ${{\left( x+1 \right)}^{2}}$; then we get
$\begin{align}
& 2{{x}^{2}}+8x+3=A{{\left( x+1 \right)}^{2}}+B\left( x+1 \right)+C \\
& =A\left( {{x}^{2}}+1+2x \right)+Bx+B+C \\
& =A{{x}^{2}}+A+2Ax+Bx+B+C \\
& =A{{x}^{2}}+\left( 2A+B \right)x+A+B+C
\end{align}$
Then, compare the coefficient of ${{x}^{2}},\ x$ and constant term:
$A=2$ …… (1)
$2A+B=8$ …… (2)
$A+B+C=3$ …… (3)
And put the value of A in equation (2):
$\begin{align}
& 2\left( 2 \right)+B=8 \\
& B=8-4
\end{align}$
$B=4$ …… (4)
Also, put the value of A and B in equation (3):
$\begin{align}
& 2+4+C=3 \\
& C=3-6 \\
& =-3
\end{align}$
Therefore,
$\frac{2{{x}^{2}}+8x+3}{{{\left( x+1 \right)}^{3}}}=\frac{2}{\left( x+1 \right)}+\frac{4}{{{\left( x+1 \right)}^{2}}}-\frac{3}{{{\left( x+1 \right)}^{3}}}$
Thus, the partial fraction of the provided expression is $\frac{2}{\left( x+1 \right)}+\frac{4}{{{\left( x+1 \right)}^{2}}}-\frac{3}{{{\left( x+1 \right)}^{3}}}$.
