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