The integral $\int e^x\left(\frac{x-1}{2 x^2}\right) dx$ is equal to:

$\frac{e^x}{2 x}+C$, where C is constant of integration

$I =\int e^x\left[\frac{x}{2 x^2}-\frac{1}{2 x^2}\right] d x$

$=\frac{1}{2} \int e^x\left(\frac{1}{x}-\frac{1}{x^2}\right) d x$

$I =\frac{1}{2} \int e^x \frac{1}{x}-e^x \frac{1}{x^2} d x$

if a function $g(x) = e^xf(x)$

then $g'(x) = e^xf(x)+e^xf'(x)$

→ so $\int e^x f(x) + e^x f'(x) dx = e^xf(x)+C$

in this case $f(x) = \frac{1}{x}$         $f'(x)=-\frac{1}{x^2}$

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