CUET UG 2025 — Mathematics Calculus

$$(x-3) = (x-1) - 2$$(x−3)=(x−1)−2

---

$$\int \frac{(x-3)e^x}{(x-1)^3} dx$$∫(x−1)3(x−3)ex​dx

$$= \int \frac{[(x-1)-2]e^x}{(x-1)^3} dx$$=∫(x−1)3[(x−1)−2]ex​dx

$$= \int \frac{(x-1)e^x}{(x-1)^3} dx - \int \frac{2e^x}{(x-1)^3} dx$$=∫(x−1)3(x−1)ex​dx−∫(x−1)32ex​dx

$$= \int \frac{e^x}{(x-1)^2} dx - 2\int \frac{e^x}{(x-1)^3} dx$$=∫(x−1)2ex​dx−2∫(x−1)3ex​dx

---

For $f(x) = \frac{1}{(x-1)^2}$f(x)=(x−1)21​:

$$f'(x) = \frac{-2}{(x-1)^3}$$f′(x)=(x−1)3−2​

Therefore:

$$f(x) + f'(x) = \frac{1}{(x-1)^2} + \frac{-2}{(x-1)^3}$$f(x)+f′(x)=(x−1)21​+(x−1)3−2​

$$= \frac{1}{(x-1)^2} - \frac{2}{(x-1)^3}$$=(x−1)21​−(x−1)32​

This matches the integrand.

---

Using $\displaystyle \int e^x[f(x) + f'(x)]dx = e^x f(x) + C$∫ex[f(x)+f′(x)]dx=exf(x)+C:

$$\int e^x\left[\frac{1}{(x-1)^2} - \frac{2}{(x-1)^3}\right]dx$$∫ex[(x−1)21​−(x−1)32​]dx

$$= e^x \cdot \frac{1}{(x-1)^2} + C$$=ex⋅(x−1)21​+C

$$= \frac{e^x}{(x-1)^2} + C$$=(x−1)2ex​+C