Answer
$$\frac{A}{x} + \frac{B}{{{x^2}}} + \frac{C}{{x - 1}}$$
Work Step by Step
$$\eqalign{
& \frac{{2x - 3}}{{{x^3} - {x^2}}} \cr
& {\text{Factoring the denominator}}{\text{, the gcf is }}{x^2} \cr
& \frac{{2x - 3}}{{{x^3} - {x^2}}} = \frac{{2x - 3}}{{{x^2}\left( {x - 1} \right)}} \cr
& {x^2}{\text{ is a linear repeated factor.}}{\text{ Its decomposition is }}\frac{A}{x} + \frac{B}{{{x^2}}} \cr
& x - 1{\text{ is a linear factor }} \cr
& {\text{Then}}{\text{,}} \cr
& \frac{{2x - 3}}{{{x^2}\left( {x - 1} \right)}} = \frac{A}{x} + \frac{B}{{{x^2}}} + \frac{C}{{x - 1}} \cr
& \frac{{2x - 3}}{{{x^3} - {x^2}}} = \frac{A}{x} + \frac{B}{{{x^2}}} + \frac{C}{{x - 1}} \cr} $$
