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

$\frac{e^x}{(x - 1)^2} + C$, $C$ is an arbitrary constant

The correct answer is Option (3) → $\frac{e^x}{(x - 1)^2} + C$, $C$ is an arbitrary constant

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

$Let \ u = x-1 \ \Rightarrow \ du = dx, \ x-3 = (u+1) - 3 = u - 2$

$I = \int \frac{(u-2)e^{u+1}}{u^3} \, du$

$= e \int \frac{u-2}{u^3} e^u \, du$

$= e \int \left(u^{-2} - 2u^{-3}\right) e^u \, du$

Use integration by parts for each term:

$\int u^{-n} e^u \, du = e^u \left( u^{-n} - n u^{-n-1} + \dots \right)$

After simplification, the result is:

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