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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.2012-03-13

1 Answers 1

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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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    @KirthiRaman of course. Thanks.2012-03-14