TITLE: Integrating $\log(x)\cdot\exp(x)$
QUESTION [10 upvotes]: 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??

REPLY [6 votes]: 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.