The value of int frac{x e^{x} d x}{(1+x)^{2}} | vedclass

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(D) Let $I = \int \frac{x e^{x} d x}{(1+x)^{2}}$.We can rewrite the numerator as $(x+1-1)$.$I = \int \frac{e^{x}(x+1-1)}{(1+x)^{2}} d x$.$I = \int e^{

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

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(D) Let $I = \int \frac{x e^{x} d x}{(1+x)^{2}}$.We can rewrite the numerator as $(x+1-1)$.$I = \int \frac{e^{x}(x+1-1)}{(1+x)^{2}} d x$.$I = \int e^{x} \left[ \frac{x+1}{(1+x)^{2}} - \frac{1}{(1+x)^{2}} \right] d x$.$I = \int e^{x} \left[ \frac{1}{1+x} - \frac{1}{(1+x)^{2}} \right] d x$.Let $f(x) = \frac{1}{1+x}$. Then $f'(x) = -\frac{1}{(1+x)^{2}}$.Using the standard integration formula $\int e^{x} [f(x) + f'(x)] d x = e^{x} f(x) + C$,we get:$I = e^{x} \left( \frac{1}{1+x} \right) + C = \frac{e^{x}}{1+x} + C$.

The value of $\int \frac{x e^{x} d x}{(1+x)^{2}}$ is equal to

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