How to integrate $\int_1^e \ln{x} \, dx$ This seems really tricky to me. I can't figure out how to integrate $\ln x$. 
 A: 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$x \mid_{1}^{e}- \int_{1}^{e} dx$
A: 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.
A: $$
\int\ln x\,dx=\int u\,dx = xu-\int x\,du = x\ln x - \int x\left(\frac1x\,dx\right).
$$
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.
A: 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.
