Answer
The partial fraction decomposition is $x+1-\frac{2}{x}-\frac{2}{{{x}^{2}}}+\frac{2}{x-1}$.
Work Step by Step
Let us consider the expression,
$\frac{{{x}^{4}}-{{x}^{2}}+2}{{{x}^{3}}-{{x}^{2}}}$
Firstly, solve by long division method,
${{x}^{3}}-{{x}^{2}}\overset{x+1}{\overline{\left){\begin{align}
& {{x}^{4}}-{{x}^{2}}+2 \\
& {{x}^{4}}\text{ }-{{x}^{3}} \\
& -\text{ }+ \\
& \overline{\begin{align}
& \text{ }{{x}^{3}}-{{x}^{2}}+2 \\
& \text{ }{{x}^{3}}-{{x}^{2}} \\
& \,\,\,-\text{ }+\text{ } \\
& \overline{\text{ }2} \\
\end{align}} \\
\end{align}}\right.}}$
So,
$\frac{{{x}^{4}}-{{x}^{2}}+2}{{{x}^{3}}-{{x}^{2}}}=x+1+\frac{2}{{{x}^{3}}-{{x}^{2}}}$
Now, partial fraction of $\frac{2}{{{x}^{3}}-{{x}^{2}}}=\frac{2}{{{x}^{2}}\left( x-1 \right)}$
And to find partial fraction decomposition,
$\frac{2}{{{x}^{2}}\left( x-1 \right)}=\frac{A}{x}+\frac{B}{{{x}^{2}}}+\frac{C}{x-1}$ …… (1)
And multiply both sides by ${{x}^{2}}\left( x-1 \right)$,
$\begin{align}
& 2=Ax\left( x-1 \right)+B\left( x-1 \right)+C{{x}^{2}} \\
& 2=A{{x}^{2}}-Ax+Bx-B+C{{x}^{2}} \\
& 2=\left( A+C \right){{x}^{2}}-\left( A-B \right)x-B \\
\end{align}$
And equate the terms,
$\begin{align}
& A+C=0 \\
& A-B=0 \\
& B=-2 \\
\end{align}$
Then, solving the above equation,
$A=-2,B=-2,C=2$
Put these values in equation (1),
$\frac{2}{{{x}^{2}}\left( x-1 \right)}=-\frac{2}{x}-\frac{2}{{{x}^{2}}}+\frac{2}{x-1}$
Thus, the partial fraction decomposition of the remainder term is $x+1-\frac{2}{x}-\frac{2}{{{x}^{2}}}+\frac{2}{x-1}$.
