Question
Evaluate: $\int \frac{e^x(x-3)}{(x-1)^3} d x$

Answer

$\int \frac{e^x(x-3)}{(x-1)^3} d x=\int e^x\left[\frac{(x-1)-2}{(x-1)^3}\right] d x$
$=\int e^x\left[\frac{1}{(x-1)^2}+\left(\frac{-2}{(x-1)^3}\right)\right] d x$
$\left(\because \int e^x\left[f(x)+f^{\prime}(x)\right] d x=e^x f(x)+c\right)$
$1=\frac{e^x}{(x-1)^2}+C$

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