Answer
$ \displaystyle \frac{4}{x(x-1)}=\frac{-4}{x}+\frac{4}{x-1}$
Work Step by Step
This is Case 1: Q has only nonrepeated linear factors.
$\displaystyle \frac{4}{x(x-1)}=\frac{A}{x}+\frac{B}{x-1} $
... write the RHs with a common denominator
$ \displaystyle \frac{4}{x(x-1)}=\frac{A(x-1)+Bx}{x(x-1)}$
... equate the numerators
$4=Ax-A+Bx$
$4=(A+B)x-A$
... equate the coefficients of the polynomials on the LHS and RHS:
$\left\{\begin{array}{lll}
0 & =A+B & \\
4 & =-A & \Rightarrow A=-4
\end{array}\right.$
Substitute $A=-4$ into the first equation
$B-4=0$
$B=4$
$ \displaystyle \frac{4}{x(x-1)}=\frac{-4}{x}+\frac{4}{x-1}$
