$\int e^x \frac{(x-1)(x-\log x)}{x^2} d x$ is equal to

$e^x\left(\frac{x-\ln x-1}{x}\right)+C$

Let $I=\int e^x \frac{(x-1)(x-\log x)}{x^2} d x$. Then,

$I=\int e^x\left(\frac{\left.x^2-x+\log x-x \log x\right.}{x^2}\right) d x$

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

$\Rightarrow I=e^x\left(\frac{x-\log x-1)}{x}\right)+C$