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

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(B) Let $I = \int \frac{(x-1) e^x}{(x+1)^3} \,d x$ We can rewrite the numerator as $(x+1-2)$: $I = \int \frac{(x+1-2) e^x}{(x+1)^3} \,d x$ $I = \int e

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

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(B) Let $I = \int \frac{(x-1) e^x}{(x+1)^3} \,d x$ We can rewrite the numerator as $(x+1-2)$: $I = \int \frac{(x+1-2) e^x}{(x+1)^3} \,d x$ $I = \int e^x \left[ \frac{x+1}{(x+1)^3} - \frac{2}{(x+1)^3} \right] d x$ $I = \int e^x \left[ \frac{1}{(x+1)^2} - \frac{2}{(x+1)^3} \right] d x$ Let $f(x) = \frac{1}{(x+1)^2} = (x+1)^{-2}$. Then $f'(x) = -2(x+1)^{-3} = -\frac{2}{(x+1)^3}$. Using the standard integral formula $\int e^x [f(x) + f'(x)] \,d x = e^x f(x) + c$: $I = e^x \left( \frac{1}{(x+1)^2} \right) + c = \frac{e^x}{(x+1)^2} + c$.

The value of $\int \frac{(x-1) e^x}{(x+1)^3} \,d x$ is equal to

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