Express $\frac{2x^3 - 5}{x^2(x - 1)}$ in partial fractions.

1. Recognise that the fraction is improper and perform long division to extract the polynomial part.

The degree of the numerator ($2x^3 - 5$) is 3, and the degree of the denominator ($x^3 - x^2$) is also 3. Thus, the fraction is improper.
$\frac{2x^3 - 5}{x^3 - x^2} = 2 + \frac{2x^2 - 5}{x^2(x - 1)}$

2. Set up the partial fraction decomposition for the proper fraction part.

Let $\frac{2x^2 - 5}{x^2(x - 1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x - 1}$
Multiplying throughout by $x^2(x - 1)$:
$2x^2 - 5 = Ax(x - 1) + B(x - 1) + Cx^2$

3. Solve for the constants $B$ and $C$ by substituting suitable values of $x$.

Substitute $x = 0$:
$2(0)^2 - 5 = A(0)(-1) + B(-1) + C(0)^2$
$-5 = -B \implies B = 5$

Substitute $x = 1$:
$2(1)^2 - 5 = A(1)(0) + B(0) + C(1)^2$
$-3 = C \implies C = -3$

4. Solve for the constant $A$ by comparing coefficients.

Compare the coefficients of $x^2$ on both sides:
$2 = A + C$
Since $C = -3$:
$2 = A - 3 \implies A = 5$

5. Write the final expression in partial fractions.

$\frac{2x^3 - 5}{x^2(x - 1)} = 2 + \frac{5}{x} + \frac{5}{x^2} - \frac{3}{x - 1}$