Answer
$\displaystyle \frac{1}{x(x-1)}=-\frac{1}{x}+\frac{1}{x-1}$
Work Step by Step
Both factors of the denominator are distinct linear factors...
see p.541. Decomposition with Distinct Linear Factors In the Denominator:
Include one partial fraction with a $constant$ numerator for each distinct linear factor in the denominator.
Setup:
$\displaystyle \frac{1}{x(x-1)}=\frac{A}{x}+\frac{B}{x-1}\qquad /\times x(x-1)\quad$(LCD)
$1=A(x-1)+Bx$
$1=Ax-A+Bx \qquad$ /group like terms...
$0\cdot x+1=x(A+B)+(-A)$
(Emphasize the 0 linear constant on the LHS,
as we now equate the coefficients of like powers...)
$\left\{\begin{array}{l}
0=A+B\\
1=-A
\end{array}\right.\qquad $solve the system
The second equation gives us: $A=-1$
back substitute:
$0=-1+B$
$B=1$
Rewrite the setup decomposition with $A=-1, B=1$
$\displaystyle \frac{1}{x(x-1)}=-\frac{1}{x}+\frac{1}{x-1}$
