Integrating $\log(x)\cdot\exp(x)$

How do I evaluate $\int \log(x) e^x\;dx ?$ I tried to do integration by parts...

$$\int\log(x) \; dx = (x-1)\log(x) $$

Let $I=\int\log(x) e^x \; dx$. Therefore, $$ I = (x-1)\log(x) e^x - \int (x-1)\log(x) e^x \; dx $$ $$= (x-1)\log(x) e^x - \int (x)\log(x) e^x \; dx +\int \log(x) e^x \; dx$$ $$= (x-1)\log(x) e^x - \int x \log(x) e^x \; dx + I$$

Now what to do, $I$ is on both LHS and RHS??

Recall $$\frac{d}{dx} e^x = e^x, \frac{d}{dx} \log x = \frac{1}{x}.$$ Now, integrate by parts $$ \int e^x \log x \ dx = e^x \log x \color{red}{-} \int \frac{e^x}{x} dx = e^x \log x \color{red}{-} \operatorname{Ei}(x) + c$$ where $\operatorname{Ei}(x)$ is Exponential Integral.

As noted in the comment by Patrick, and this paragraph in Wikipedia:

$\int \frac{e^x}{x} dx $ is not an elementary function, a fact which can be proven using the Risch Algorithm.