The value of I = int frac{(x-1) e^x}{(x+1)^3} | vedclass

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(D) $I = \int \frac{(x-1) e^x}{(x+1)^3} \,dx$ $I = \int \left( \frac{x+1-2}{(x+1)^3} \right) e^x \,dx$ $I = \int \left[ \frac{x+1}{(x+1)^3} - \frac{2}

Vedclass · Accepted Answer

$\frac{e^x}{(x+1)^2} + C$,(where $C$ is a constant of integration)

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(D) $I = \int \frac{(x-1) e^x}{(x+1)^3} \,dx$ $I = \int \left( \frac{x+1-2}{(x+1)^3} \right) e^x \,dx$ $I = \int \left[ \frac{x+1}{(x+1)^3} - \frac{2}{(x+1)^3} \right] e^x \,dx$ $I = \int \left[ \frac{1}{(x+1)^2} - \frac{2}{(x+1)^3} \right] e^x \,dx$ Let $f(x) = \frac{1}{(x+1)^2}$,then $f'(x) = -2(x+1)^{-3} = \frac{-2}{(x+1)^3}$. Using the formula $\int e^x (f(x) + f'(x)) \,dx = e^x f(x) + C$,we get: $I = e^x \left( \frac{1}{(x+1)^2} \right) + C = \frac{e^x}{(x+1)^2} + C$.

The value of $I = \int \frac{(x-1) e^x}{(x+1)^3} \,dx$ is

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