# How to integrate $\int_1^e \ln{x} \, dx$

This seems really tricky to me. I can't figure out how to integrate $\ln x$.

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Try to look it up in some integration table. That would help. – Hui Yu Jan 16 at 16:03
Integrate $1\cdot\ln x$ by parts ;) – CBenni Jan 16 at 16:03
@HuiYu not if this is homework or some other assignment – CBenni Jan 16 at 16:05
@CBenni Oh, yeah... – Hui Yu Jan 16 at 16:11
Nah, I had it in my test. :-( – weasel Jan 16 at 16:14

Based upon geometrical observation we write that

$$\int_1^e \ln x\space \mathrm{dx}=e-\int_0^1 e^x\space \mathrm{dx}=1$$

The question may also be viewed as a particular case of Young's inequality.

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That is MUCH nicer (IMHO) then the regular integration by parts trick! – Amihai Zivan Jan 16 at 16:21
@AmihaiZivan: thanks! :-) – Chris's wise sister Jan 16 at 16:22
@Chris'ssister I wonder what did you do here? Can you please explain? – weasel Jan 16 at 16:44
@weasel: I simply used the inverse function of $\ln x$. Make a drawing and see what happens in the rectangle $[1, e] \times [0, 1]$ if you employ the inverse function. – Chris's wise sister Jan 16 at 16:47
Thank you. I learned a new method, here. – weasel Jan 22 at 15:18

Hint Integration by parts is helpful.

Let $u=$ln$x$ , $dv=1dx$
so $du=\frac{1}{x} dx$ , $v=x$

so that $uv$ - $\int_{1}^{e}vdu$ = $x$ln$x \mid_{1}^{e}- \int_{1}^{e} dx$

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$$\int\ln x\,dx=\int u\,dx = xu-\int x\,du = x\ln x - \int x\left(\frac1x\,dx\right).$$ Now here's the hard part: $x\cdot\dfrac1x$ simplifies to $1$. At least, I've seen lots of students get stuck on that part. Some of them want to antidifferentiate $x$ and also $\dfrac1x$.

So you've got $$x\ln x-\int 1\,dx.$$ You can probably do the rest.

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 (+1) I remember somewhere some answer was upvoted for the humour in answer. – 007resu Jan 16 at 18:23

Change variable formula $\int_{\varphi(a)}^{\varphi(b)}f (x) \, dx = \int_a^b f(\varphi(u))\varphi^\prime (u) du$ \begin{align} \int_{1}^{e}\ln (x)\, dx = & \int_{e^{0}}^{e^{1}}\ln (x) \, dx \\ = & \int_0^1\ln( e^u ) (e^u)^\prime du \\ = & \int_0^1 u \cdot e^u du \\ \end{align} Now use integration by parts.

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 And what is u here? – weasel Jan 17 at 3:26