int frac{x e^x}{(1 + x)^2} dx = | vedclass

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(C) We need to evaluate the integral $I = \int \frac{x e^x}{(1 + x)^2} dx$. Rewrite the numerator as $(x + 1 - 1)$: $I = \int \frac{(x + 1 - 1) e^x}{(

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$\frac{e^x}{1 + x} + c$

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(C) We need to evaluate the integral $I = \int \frac{x e^x}{(1 + x)^2} dx$. Rewrite the numerator as $(x + 1 - 1)$: $I = \int \frac{(x + 1 - 1) e^x}{(1 + x)^2} dx$ Split the fraction: $I = \int e^x \left( \frac{x + 1}{(1 + x)^2} - \frac{1}{(1 + x)^2} \right) dx$ $I = \int e^x \left( \frac{1}{1 + x} - \frac{1}{(1 + x)^2} \right) dx$ Recall the standard integral formula $\int e^x [f(x) + f'(x)] dx = e^x f(x) + c$. Here,let $f(x) = \frac{1}{1 + x}$. Then $f'(x) = -\frac{1}{(1 + x)^2}$. Thus,the integral becomes: $I = e^x \left( \frac{1}{1 + x} \right) + c = \frac{e^x}{1 + x} + c$.

$\int \frac{x e^x}{(1 + x)^2} dx = $

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