Integrate the function: $\frac{(x-3) e^{x}}{(x-1)^{3}}$

We need to evaluate the integral $I = \int \frac{(x-3) e^{x}}{(x-1)^{3}} dx$.
Rewrite the numerator as $(x-1-2)$:
$I = \int e^{x} \left\{ \frac{x-1-2}{(x-1)^{3}} \right\} dx$
Split the fraction:
$I = \int e^{x} \left\{ \frac{x-1}{(x-1)^{3}} - \frac{2}{(x-1)^{3}} \right\} dx = \int e^{x} \left\{ \frac{1}{(x-1)^{2}} - \frac{2}{(x-1)^{3}} \right\} dx$
Let $f(x) = \frac{1}{(x-1)^{2}}$. Then,$f'(x) = -2(x-1)^{-3} = \frac{-2}{(x-1)^{3}}$.
Using the standard integral formula $\int e^{x} \{f(x) + f'(x)\} dx = e^{x} f(x) + C$:
$I = e^{x} \left( \frac{1}{(x-1)^{2}} \right) + C = \frac{e^{x}}{(x-1)^{2}} + C$,where $C$ is an arbitrary constant.