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

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(B) We know that $\int e^x [f(x) + f'(x)] \, dx = e^x f(x) + c$. Given integral is $I = \int e^x \frac{x - 3}{(x - 1)^3} \, dx$. Rewrite the numerator

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) We know that $\int e^x [f(x) + f'(x)] \, dx = e^x f(x) + c$. Given integral is $I = \int e^x \frac{x - 3}{(x - 1)^3} \, dx$. Rewrite the numerator as $(x - 1) - 2$: $I = \int e^x \frac{(x - 1) - 2}{(x - 1)^3} \, dx$ $I = \int e^x \left[ \frac{x - 1}{(x - 1)^3} - \frac{2}{(x - 1)^3} \right] \, dx$ $I = \int e^x \left[ \frac{1}{(x - 1)^2} - \frac{2}{(x - 1)^3} \right] \, dx$ 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 integral is in the form $\int e^x [f(x) + f'(x)] \, dx$,the result is $e^x f(x) + c$. Therefore,$I = \frac{e^x}{(x - 1)^2} + c$.

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

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