Answer
$\frac{-2x^2+5x-1}{(x-1)^3(x+1)}=\frac{1}{x+2}+\frac{-1}{x-1}+\frac{1}{(x-1)^3}$,
Work Step by Step
$\frac{-2x^2+5x-1}{x^4-2x^3+2x-1}$ = factorising $x^4-2x^3+2x-1=(x^2-2x+1)(x^2-1)=(x-1)(x+1)(x-1)(x-1)=(x-1)^3(x+1)$,
$\frac{-2x^2+5x-1}{(x+1)(x-1)^3}=\frac{A}{x+1}+\frac{B}{x-1}+\frac{C}{(x-1)^2}+\frac{D}{(x-1)^3}$,
$A(x-1)^3+B(x-1)^2(x+1)+C(x-1)(x+1)+D(x+1)$,
$A(x^3-3x^2+3x-1)+B(x^3-x^2-x+1)+C(x^2-1)+D(x+1)$,
$(A+B)x^3+(-3A-B+C)x^2+(3A-B+D)x+(-A+B-C+D)$,
$\begin{cases}
& A & +B=0\\
& -3A & -B +C = -2\\
3A & -B & +D=5\\
-A & +B & -C +D= -1
\end{cases}$
Adding Equation $2$ and Equation $3$.
$\begin{cases}
-3A -B +C=-2\\
3A -B +D=5\\
-- -- -- -- -- -- -- --\\
0A -2B +C +D=3
\end{cases}$
Adding Equation $1$ and Equation $4$.
$\begin{cases}
& A +B=0\\
& -A +B -C +D=-1\\
& -- -- -- -- -- -- -- --\\
& 0A +2B -C +D=-1
\end{cases}$
Adding the Addition result of Equation $2$, Equation $3$ and the Addition result of Equation $1$, Equation $4$
$\begin{cases}
& 2B -C +D=-1\\
& -2B +C +D=3\\
& -- -- -- -- -- --\\
& 0C +2D=2
\end{cases}$
thus, $D=1$,
Adding Equation $1$ and Equation $3$.
$\begin{cases}
& A +B=0\\
& 3A-B=4\\
& -- -- -- -- --\\
& 4A +0B=4
\end{cases}$
thus, $A=1$ Substituting back into the Equation, $1+B=0, B=-1$.
Substituting back into the Equation, $-3+1+C=-2, C=0$.
Therefore,
$\frac{-2x^2+5x-1}{(x-1)^3(x+1)}=\frac{1}{x+2}+\frac{-1}{x-1}+\frac{1}{(x-1)^3}$,
