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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??

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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
up vote 4 down vote accepted

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.

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In case you don't notice, J.D.'s answer means that if no one has a closed formula for the exponential integral, then no one has a closed formula for the integral you are looking for. This means that you can't hope (I believe) to "compute it" as you wished. – Patrick Da Silva Mar 13 '12 at 2:20
I believe it is $e^x \log x - \operatorname{Ei}(x) + c$ – Kirthi Raman Mar 14 '12 at 15:39
@KirthiRaman of course. Thanks. – user2468 Mar 14 '12 at 16:34

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