Answer
$\dfrac{A}{x}+\dfrac{B}{x^2}+\dfrac{C}{x^3}+\dfrac{D}{x-1}$
Work Step by Step
Simplify $\dfrac{1}{x^4-x^3}$ into partial fraction decomposition.
This can be written as:
$\dfrac{1}{x^4-x^3}=\dfrac{1}{x^3(x-1)}$
Applying the partial fraction decomposition rule.
$\dfrac{1}{x^3(x-1)}=\dfrac{A}{x}+\dfrac{B}{x^2}+\dfrac{C}{x^3}+\dfrac{D}{x-1}$
