Answer
$\dfrac{1}{x^2-ax-bx+ab}=-\dfrac{1}{(b-a)(x-a)}+\dfrac{1}{(b-a)(x-b)}$
Work Step by Step
We are given the fraction:
$\dfrac{1}{x^2-ax-bx+ab}$
Factor the denominator:
$\dfrac{1}{x^2-ax-bx+ab}=\dfrac{1}{x(x-a)-b(x-a)}=\dfrac{1}{(x-a)(x-b)}$
Write the partial fraction decomposition:
$\dfrac{1}{(x-a)(x-b)}=\dfrac{A}{x-a}+\dfrac{B}{x-b}$
Multiply all terms by the least common denominator $(x-a)(x-b)$:
$(x-a)(x-b)\cdot\dfrac{1}{(x-a)(x-b)}=(x-a)(x-b)\cdot\dfrac{A}{x-a}+(x-a)(x-b)\cdot\dfrac{B}{x-b}$
$1=A(x-b)+B(x-a)$
$1=Ax-Ab+Bx-Ba$
$1=(A+B)x+(-Ab-Ba)$
Identify the coefficients:
$\begin{cases}
A+B=0\\
-Ab-Ba=1
\end{cases}$
$\begin{cases}
Ab+Bb=0(b)\\
-Ab-Ba=1
\end{cases}$
$Ab+Bb-Ab-Ba=0+1$
$B(b-a)=1$
$B=\dfrac{1}{b-a}$
$A+B=0$
$A=-B$
$A=-\dfrac{1}{b-a}$
The partial fraction decomposition is:
$\dfrac{1}{(x-a)(x-b)}=-\dfrac{\dfrac{1}{b-a}}{x-a}+\dfrac{\dfrac{1}{b-a}}{(x-b)^2}$
$\dfrac{1}{x^2-ax-bx+ab}=-\dfrac{1}{(b-a)(x-a)}+\dfrac{1}{(b-a)(x-b)}$
