# 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??

-
It is, but you made your integral more complicated... so I don't think this is the way to go. –  Patrick Da Silva Mar 13 '12 at 2:17

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.
$\int \frac{e^x}{x} dx$ is not an elementary function, a fact which can be proven using the Risch Algorithm.
I believe it is $e^x \log x - \operatorname{Ei}(x) + c$ –  Kirthi Raman Mar 14 '12 at 15:39