Definite integral involving $\sqrt{\log}$ In some old lecture notes I found the following problem:
$$ \int_1^e \frac{1+2x \sqrt{\log x}}{2 x \sqrt{\log  x}(x+\sqrt{\log x})} \;dx$$
I just don't seem to get a handle on this rather difficult looking integral. Does anybody have some insights?

First note that
\begin{align*}
\left(\sqrt{\log{x}}\right)'=\frac{1}{2x\sqrt{\log{x}}}
\end{align*}
Then for the integrand it follows that
\begin{align*}
\frac{1+2x\sqrt{\log x}}{2x\sqrt{\log x}(x+\sqrt{\log x})} &=\frac{\frac{1}{2x\sqrt{\log x}}+1}{x+\sqrt{\log x}}\\
&=\frac{(x+\sqrt{\log x})'}{x+\sqrt{\log x}}.
\end{align*} 
This enables us to calculate
\begin{align*}
\int\frac{(x+\sqrt{\log x})'}{x+\sqrt{\log x}}dx=\log\left(x+\sqrt{\log x}\right)
\end{align*}
and therefore
\begin{align*}
\int^e_1\frac{(x+\sqrt{\log x})'}{x+\sqrt{\log x}}dx =\log\left(e+\sqrt{\log e}\right)-\log\left(1+\sqrt{\log 1}\right)=\log\left(e+1\right).
\end{align*}
