int e^x frac{(x-1)}{(x+1)^3} , dx = | vedclass

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(C) We want to evaluate the integral $I = \int e^x \frac{x-1}{(x+1)^3} \, dx$. Rewrite the numerator as $(x+1) - 2$: $I = \int e^x \frac{(x+1) - 2}{(x

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$\frac{e^x}{(x+1)^2}+c$

Vedclass · Answer

(C) We want to evaluate the integral $I = \int e^x \frac{x-1}{(x+1)^3} \, dx$. Rewrite the numerator as $(x+1) - 2$: $I = \int e^x \frac{(x+1) - 2}{(x+1)^3} \, dx$ $I = \int e^x \left( \frac{x+1}{(x+1)^3} - \frac{2}{(x+1)^3} \right) \, dx$ $I = \int e^x \left( \frac{1}{(x+1)^2} - \frac{2}{(x+1)^3} \right) \, dx$ Recall the standard integral form $\int e^x [f(x) + f'(x)] \, dx = e^x f(x) + c$. Let $f(x) = \frac{1}{(x+1)^2} = (x+1)^{-2}$. Then $f'(x) = -2(x+1)^{-3} = -\frac{2}{(x+1)^3}$. Since the integrand is in the form $e^x [f(x) + f'(x)]$,the integral is $e^x f(x) + c$. Therefore,$I = \frac{e^x}{(x+1)^2} + c$.

$\int e^x \frac{(x-1)}{(x+1)^3} \, dx =$

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