CUET UG 2025 — Mathematics Calculus

Given: $\displaystyle \int \frac{(x-1)e^x}{x^2} dx, x > 0$∫x2(x−1)ex​dx,x>0

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Consider $\frac{e^x}{x}$xex​ and differentiate using the Quotient Rule:

$$\frac{d}{dx}\left(\frac{e^x}{x}\right) = \frac{x \cdot \frac{d}{dx}(e^x) - e^x \cdot \frac{d}{dx}(x)}{x^2}$$dxd​(xex​)=x2x⋅dxd​(ex)−ex⋅dxd​(x)​

$$= \frac{x \cdot e^x - e^x \cdot 1}{x^2}$$=x2x⋅ex−ex⋅1​

$$= \frac{xe^x - e^x}{x^2}$$=x2xex−ex​

$$= \frac{e^x(x-1)}{x^2}$$=x2ex(x−1)​

$$= \frac{(x-1)e^x}{x^2}$$=x2(x−1)ex​

This matches the original integrand.

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Since $\frac{d}{dx}\left(\frac{e^x}{x}\right) = \frac{(x-1)e^x}{x^2}$dxd​(xex​)=x2(x−1)ex​

$$\int \frac{(x-1)e^x}{x^2} dx = \frac{e^x}{x} + C$$∫x2(x−1)ex​dx=xex​+C