Answer
$\frac{5x^2-3x+10}{(x^2+2)(x-1)(x+1)}=\frac{2}{x-1}+\frac{-3}{x+1}+\frac{x}{x^2+2}$
Work Step by Step
$\frac{5x^2-3x+10}{x^4+x^2-2}$, factorising $x^4+x^2-2=(x^2-1)(x^2+2)=(x-1)(x+1)(x^2+2)$,
thus,
$\frac{5x^2-3x+10}{(x-1)(x+1)(x^2+2)}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{Cx+D}{x^2+2}$,
$A(x+1)(x^2+2)+B(x-1)(x^2+2)+(Cx+D)(x+1)(x-1)$,
$(A+B+C)x^3+(A-B+D)x^2+(2A+2B-C)x+(2A-2B-D)=5x^2-3x+10$,
$\begin{cases}
A+B+C=0\\
A-B+D=5\\
2A+2B-C=-3\\
2A-2B-D=10
\end{cases}$,
Adding Equation 1 and Equation 2.
$\begin{cases}
A+B+C=0\\
A-B+D=5\\
-- -- -- --\\
2A+C+D=5
\end{cases}$
Adding Equation 3 and Equation 4.
$\begin{cases}
2A+2B-C=-3\\
2A-2B-D=10\\
-- -- -- --\\
4A-C-D=7
\end{cases}$
$\begin{cases}
2A+C+D=5\\
4A-C-D=7\\
-- -- -- -- -- -\\
6A=12
\end{cases}$
thus, $A=2$, substituting back into the Equation 1 $B+C=-2$ and Equation 3 $2B-C=-7$.
Adding Equation 1 and Equation 3.
$\begin{cases}
B+C=-2\\
2B-C=-7\\
-- -- -- --\\
3B=-9
\end{cases}$
thus, $B=-3$, substituting into the Equation, $-3+C=-2, C=1$ and $5+D=5, D=0$.
Therefore,
$\frac{5x^2-3x+10}{(x^2+2)(x-1)(x+1)}=\frac{2}{x-1}+\frac{-3}{x+1}+\frac{x}{x^2+2}$
