Answer
$\frac{4x-1}{(x-1)^2(x+2)}=\frac{-1}{x+2}+\frac{1}{x-1}+\frac{1}{(x-1)^2}$
Work Step by Step
$\frac{4x-1}{(x-1)^2(x+2)}=\frac{A}{x+2}+\frac{B}{x-1}+\frac{C}{(x-1)^2}$,
$A(x-1)^2+B(x-1)(x+2)+C(x+2)$,
$A(x^2-2x+1)+B(x^2+x-2)+C(x+2)$,
$(A+B)x^2+(-2A+B+C)x+(A-2B+2C)=4x-1$,
$\begin{cases}
A+B=0\\
-2A+B+C=4\\
A-2B+2C=-1
\end{cases}$,
Multiplying Equation 3 by 2 and adding it to Equation 2.
$\begin{cases}
-2A+B+C=4\\
2A-4B+4C=-2\\
-- -- -- --\\
-3B+5C=2
\end{cases}$
Multiplying Equation 1 by -1 and adding it to Equation 3.
$\begin{cases}
-A-B=0\\
A-2B+2C=-1\\
-- -- -- --\\
-3B+2C=-1
\end{cases}$
Multiplying one of the addition results by -1 and adding it to another.
$\begin{cases}
-3B+5C=2\\
3B-2C=1\\
-- -- -- --\\
3C=3
\end{cases}$,
thus, $C=1$, substituting back into the Equation, $-3B+5=2, B=1$,
substituting back into the Equation, $A+1=0, A=-1$.
Therefore,
$\frac{4x-1}{(x-1)^2(x+2)}=\frac{-1}{x+2}+\frac{1}{x-1}+\frac{1}{(x-1)^2}$
