Answer
The partial fraction decomposition is ${{x}^{2}}+3x+1+\frac{3}{x+1}+\frac{5}{x-2}$.
Work Step by Step
Let us consider the expression,
$\frac{{{x}^{4}}+2{{x}^{3}}-4{{x}^{2}}+x-3}{{{x}^{2}}-x-2}$
Firstly, solve by long division method,
${{x}^{2}}-x-2\overset{{{x}^{2}}+3x+1}{\overline{\left){\begin{align}
& {{x}^{4}}+2{{x}^{3}}-4{{x}^{2}}+x-3 \\
& {{x}^{4}}-{{x}^{3}}\text{ }-2{{x}^{2}} \\
& -\text{ + +} \\
& \overline{\begin{align}
& \text{ 3}{{x}^{3}}\text{ }-\text{2}{{x}^{2}}+x-3 \\
& \text{ 3}{{x}^{3}}\text{ }-3{{x}^{2}}-\text{6}x \\
& \text{ }-\text{ + +} \\
& \overline{\begin{align}
& \text{ }{{x}^{2}}\text{ +7}x-3 \\
& \text{ }{{x}^{2}}\text{ }-x\text{ }-\text{2} \\
& \text{ }-\text{ + +} \\
& \overline{\text{ 8}x-1} \\
\end{align}} \\
\end{align}} \\
\end{align}}\right.}}$
So,
$\frac{{{x}^{4}}+2{{x}^{3}}-4{{x}^{2}}+x-3}{{{x}^{2}}-x-2}={{x}^{2}}+3x+1+\frac{8x-1}{{{x}^{2}}-x-2}$
Then, partial fraction of $\frac{8x-1}{{{x}^{2}}-x-2}=\frac{8x-1}{\left( x+1 \right)\left( x-2 \right)}$,
And to find partial fraction decomposition,
$\frac{8x-1}{\left( x+1 \right)\left( x-2 \right)}=\frac{A}{x+1}+\frac{B}{x-2}$ …… (1)
And multiply both sides by $\left( x+1 \right)\left( x-2 \right)$,
$\begin{align}
& 8x-1=A\left( x-2 \right)+B\left( x+1 \right) \\
& 8x-1=Ax-2A+Bx+B \\
& 8x-1=\left( A+B \right)x-\left( 2A-B \right) \\
\end{align}$
And equate the terms,
$\begin{align}
& A+B=8 \\
& -2A+B=-1 \\
\end{align}$
Then solving above equation,
$A=3,B=5$
Put these values in equation (1),
$\frac{8x-1}{\left( x+1 \right)\left( x-2 \right)}=\frac{3}{x+1}+\frac{5}{x-2}$
Thus, the partial fraction decomposition of the remainder term is,
${{x}^{2}}+3x+1+\frac{3}{x+1}+\frac{5}{x-2}$.
