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

$(x-3) = (x-1) - 2$

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

$= \int \frac{[(x-1)-2]e^x}{(x-1)^3} dx$

$= \int \frac{(x-1)e^x}{(x-1)^3} dx - \int \frac{2e^x}{(x-1)^3} dx$

$= \int \frac{e^x}{(x-1)^2} dx - 2\int \frac{e^x}{(x-1)^3} dx$

For $f(x) = \frac{1}{(x-1)^2}$:

$f'(x) = \frac{-2}{(x-1)^3}$

Therefore:

$f(x) + f'(x) = \frac{1}{(x-1)^2} + \frac{-2}{(x-1)^3}$

$= \frac{1}{(x-1)^2} - \frac{2}{(x-1)^3}$

This matches the integrand.

Using $\int e^x[f(x) + f'(x)]dx = e^x f(x) + C$:

$\int e^x\left[\frac{1}{(x-1)^2} - \frac{2}{(x-1)^3}\right]dx$

$= e^x \cdot \frac{1}{(x-1)^2} + C$

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