Answer
$\displaystyle \frac{e^{x}}{x}+C$
Work Step by Step
$I=\displaystyle \int\frac{(x-1)e^{x}}{x^{2}}dx=\int(x-1)e^{x}\cdot(\frac{1}{x^{2}})dx=\int udv$
(by parts)
$\left[\begin{array}{ll}
u=(x-1)e^{x} & dv =\dfrac{1}{x^{2}}dx\\
du=[(x-1)e^{x}+e^{x}]dx \\
du=[xe^{x}-e^{x}+e^{x}]dx & \\
du=xe^{x}dx & v=-\dfrac{1}{x}
\end{array}\right]$
$I=uv-\displaystyle \int vdu$
$=(x-1)e^{x}(-\displaystyle \frac{1}{x})-\int xe^{x}(-\frac{1}{x})dx$
This integral is much simpler, so we proceed.
$=(x-1)e^{x}(-\displaystyle \frac{1}{x})-\int-e^{x}dx$
$=(-1+\displaystyle \frac{1}{x})e^{x}+\int e^{x}dx$
$=-e^{x}+\displaystyle \frac{e^{x}}{x}+e^{x}+C$
$=\displaystyle \frac{e^{x}}{x}+C$
