int frac{(x-3) e^x}{(x-1)^3} d x= . . . . . | vedclass

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

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

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(D) We want to evaluate the integral $I = \int \frac{(x-3) e^x}{(x-1)^3} d x$. Rewrite the numerator as $(x-1-2)$: $I = \int \frac{(x-1-2) e^x}{(x-1)^3} d x$ $I = \int \left( \frac{x-1}{(x-1)^3} - \frac{2}{(x-1)^3} \right) e^x d x$ $I = \int \left( \frac{1}{(x-1)^2} - \frac{2}{(x-1)^3} \right) e^x d x$ Recall the standard integral form $\int e^x [f(x) + f'(x)] d x = e^x f(x) + C$. 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,$I = \frac{e^x}{(x-1)^2} + C$.

$\int \frac{(x-3) e^x}{(x-1)^3} d x=$ . . . . . . $+C$.

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