CUET UG 2025 — Mathematics Calculus

The integral to solve is $\displaystyle \int e^x \left(\frac{1-x}{1+x^2}\right)^2 dx$∫ex(1+x21−x​)2dx

For integrals with $e^x$ex, the formula $\displaystyle \int e^x [f(x) + f'(x)] dx = e^x \cdot f(x) + C$∫ex[f(x)+f′(x)]dx=ex⋅f(x)+C can be applied.

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Expanding the square:

$$\left(\frac{1-x}{1+x^2}\right)^2 = \frac{(1-x)^2}{(1+x^2)^2}$$(1+x21−x​)2=(1+x2)2(1−x)2​

The integral becomes:

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

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Let $f(x) = \frac{1}{1+x^2}$f(x)=1+x21​

Finding the derivative:

$$f'(x) = \frac{-2x}{(1+x^2)^2}$$f′(x)=(1+x2)2−2x​

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Adding $f(x)$f(x) and $f'(x)$f′(x):

$$f(x) + f'(x) = \frac{1}{1+x^2} + \frac{-2x}{(1+x^2)^2}$$f(x)+f′(x)=1+x21​+(1+x2)2−2x​

$$= \frac{(1+x^2)}{(1+x^2)^2} + \frac{-2x}{(1+x^2)^2}$$=(1+x2)2(1+x2)​+(1+x2)2−2x​

$$= \frac{1+x^2-2x}{(1+x^2)^2}$$=(1+x2)21+x2−2x​

$$= \frac{1-2x+x^2}{(1+x^2)^2}$$=(1+x2)21−2x+x2​

$$= \frac{(1-x)^2}{(1+x^2)^2}$$=(1+x2)2(1−x)2​

This matches the integrand.

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Since $\frac{(1-x)^2}{(1+x^2)^2} = f(x) + f'(x)$(1+x2)2(1−x)2​=f(x)+f′(x) where $f(x) = \frac{1}{1+x^2}$f(x)=1+x21​:

$$\int e^x \left[\frac{(1-x)^2}{(1+x^2)^2}\right] dx = e^x \cdot f(x) + C$$∫ex[(1+x2)2(1−x)2​]dx=ex⋅f(x)+C

$$= \frac{e^x}{1+x^2} + C$$=1+x2ex​+C