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

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

$\text{Let } t = x-1 \Rightarrow x = t+1$

$= \int \frac{(t-2)e^{t+1}}{t^3} dt$

$= e \int e^t \left(\frac{1}{t^2} - \frac{2}{t^3}\right) dt$

$\text{Let } u = \frac{e^t}{t^2}$

$\frac{du}{dt} = e^t\left(\frac{1}{t^2} - \frac{2}{t^3}\right)$

$ \int e^t\left(\frac{1}{t^2} - \frac{2}{t^3}\right) dt = \frac{e^t}{t^2}$

$= e \cdot \frac{e^t}{t^2} + C$

$= \frac{e^{t+1}}{t^2} + C$

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

The value is $\frac{e^x}{(x-1)^2} + C$.