∫((x-1)e^x)/(x^2) dx, x gt 0 equals (where C is an arbitrary constant)

The correct answer is Option (2) → \(\displaystyle\frac{e^x}{x}+C\)

Splitting the fraction:

    \( \displaystyle \frac{(x-1)e^x}{x^2} = e^x\left(\frac{1}{x} - \frac{1}{x^2}\right) \)

This has the form \(e^x\left[f(x) + f'(x)\right]\) with \(f(x) = \frac{1}{x}\), since \(f'(x) = -\frac{1}{x^2}\). By the standard result \(\int e^x\left[f(x) + f'(x)\right]dx = e^x f(x) + C\):

    \( \displaystyle \int \frac{(x-1)e^x}{x^2}\,dx = \frac{e^x}{x} + C \)