CUET UG 2025 — Mathematics Calculus

$$\int \frac{x^3 - 1}{x^2} dx = \int \left(\frac{x^3}{x^2} - \frac{1}{x^2}\right) dx$$∫x2x3−1​dx=∫(x2x3​−x21​)dx

---

$$= \int \left(x^{3-2} - \frac{1}{x^2}\right) dx$$=∫(x3−2−x21​)dx

$$= \int \left(x - \frac{1}{x^2}\right) dx$$=∫(x−x21​)dx

$$= \int (x - x^{-2}) dx$$=∫(x−x−2)dx

---

Using the power rule $\displaystyle \int x^n dx = \frac{x^{n+1}}{n+1} + c$∫xndx=n+1xn+1​+c:

$$\int x dx = \frac{x^{2}}{2}$$∫xdx=2x2​

$$\int x^{-2} dx = \frac{x^{-1}}{-1} = -x^{-1} = -\frac{1}{x}$$∫x−2dx=−1x−1​=−x−1=−x1​

---

$$\int \frac{x^3 - 1}{x^2} dx = \frac{x^2}{2} - \left(-\frac{1}{x}\right) + c$$∫x2x3−1​dx=2x2​−(−x1​)+c

$$= \frac{x^2}{2} + \frac{1}{x} + c$$=2x2​+x1​+c

Therefore, the answer is $\frac{x^2}{2} + \frac{1}{x} + c$2x2​+x1​+c where $c$c is the constant of integration.