Question:
$\frac{(x-3) e^{x}}{(x-1)^{3}}$
Solution:
$\int e^{x}\left\{\frac{x-3}{(x-1)^{3}}\right\} d x=\int e^{x}\left\{\frac{x-1-2}{(x-1)^{3}}\right\} d x$
$=\int e^{x}\left\{\frac{1}{(x-1)^{2}}-\frac{2}{(x-1)^{3}}\right\} d x$
Let $f(x)=\frac{1}{(x-1)^{2}} \Rightarrow f^{\prime}(x)=\frac{-2}{(x-1)^{3}}$
It is known that, $\int e^{x}\left\{f(x)+f^{\prime}(x)\right\} d x=e^{x} f(x)+\mathrm{C}$
$\therefore \int e^{x}\left\{\frac{(x-3)}{(x-1)^{2}}\right\} d x=\frac{e^{x}}{(x-1)^{2}}+C$