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

Question

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

Answer 1

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.

Answer 2

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$x \mid_{1}^{e}- \int_{1}^{e} dx$

Answer 3

$$ \int\ln x\,dx=\int u\,dx = xu-\int x\,du = x\ln x - \int x\left(\frac1x\,dx\right). $$ 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.

Answer 4

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.