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

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

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

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(A) We need to evaluate the integral $I = \int \frac{x-3}{(x-1)^3} e^x \, dx$. Rewrite the numerator as $(x-1) - 2$: $I = \int \frac{(x-1) - 2}{(x-1)^3} e^x \, dx$ $I = \int \left[ \frac{x-1}{(x-1)^3} - \frac{2}{(x-1)^3} \right] e^x \, dx$ $I = \int e^x \left[ \frac{1}{(x-1)^2} - \frac{2}{(x-1)^3} \right] \, dx$ 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 $\int e^x [f(x) + f'(x)] \, dx = e^x f(x) + c$,we get: $I = e^x \left( \frac{1}{(x-1)^2} \right) + c$.

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

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