int_0^1 frac{x e^x}{(x+1)^2} d x is equal to | vedclass

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(B) $I = \int_0^1 \frac{x e^x}{(x+1)^2} d x$ We can rewrite the integrand as: $I = \int_0^1 e^x \left[ \frac{x+1-1}{(x+1)^2} \right] d x$ $I = \int_0^

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$\frac{e}{2}-1$

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(B) $I = \int_0^1 \frac{x e^x}{(x+1)^2} d x$ We can rewrite the integrand as: $I = \int_0^1 e^x \left[ \frac{x+1-1}{(x+1)^2} \right] d x$ $I = \int_0^1 e^x \left[ \frac{1}{x+1} - \frac{1}{(x+1)^2} \right] d x$ Using the standard integral formula $\int e^x [f(x) + f'(x)] d x = e^x f(x) + C$,let $f(x) = \frac{1}{x+1}$. Then,$f'(x) = -\frac{1}{(x+1)^2}$. Thus,the integral becomes: $I = \left[ e^x \cdot \frac{1}{x+1} \right]_0^1$ $I = \left( \frac{e^1}{1+1} \right) - \left( \frac{e^0}{0+1} \right)$ $I = \frac{e}{2} - 1$

$\int_0^1 \frac{x e^x}{(x+1)^2} d x$ is equal to

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