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∫ex(1−x1+x2)2dx=\int e^x \left(\frac{1-x}{1+x^2}\right)^2 dx =

Solution

✅ Correct Option: 1

Expand: (1−x)2(1+x2)2=1−2x+x2(1+x2)2=1+x2(1+x2)2−2x(1+x2)2=11+x2−2x(1+x2)2\frac{(1-x)^2}{(1+x^2)^2} = \frac{1 - 2x + x^2}{(1+x^2)^2} = \frac{1+x^2}{(1+x^2)^2} - \frac{2x}{(1+x^2)^2} = \frac{1}{1+x^2} - \frac{2x}{(1+x^2)^2}. Let f(x)=11+x2f(x) = \frac{1}{1+x^2}, then f′(x)=−2x(1+x2)2f'(x) = -\frac{2x}{(1+x^2)^2}. Using ∫ex[f(x)+f′(x)]dx=exf(x)+C\int e^x [f(x)+f'(x)] dx = e^x f(x) + C, the answer is ex1+x2+C\frac{e^x}{1+x^2} + C.

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