Evaluate the integral: $\int \frac{x^3 - 1}{x^2} \, dx$

$\frac{x^2}{2} + \frac{1}{x} + C$

The correct answer is Option (3) → $\frac{x^2}{2} + \frac{1}{x} + C$

We have

$\int \frac{x^3 - 1}{x^2} \, dx = \int x \, dx - \int x^{-2} \, dx \quad \text{(by Property V)}$

$= \left( \frac{x^{1+1}}{1+1} + C_1 \right) - \left( \frac{x^{-2+1}}{-2+1} + C_2 \right); \quad C_1, C_2 \text{ are constants of integration}$

$= \frac{x^2}{2} + C_1 - \frac{x^{-1}}{-1} - C_2 = \frac{x^2}{2} + \frac{1}{x} + C_1 - C_2 \text{}$

$= \frac{x^2}{2} + \frac{1}{x} + C, \text{ where } C = C_1 - C_2 \text{ is another constant of integration.}$