\begin{eqnarray*}
\int_{0}^{1}{x - 1 \over \ln\left(x\right)}\,{\rm d}x
& = &
\int_{0}^{\infty}{{\rm e}^{-x} - {\rm e}^{-2x} \over x}\,{\rm d}x
=
-\int_{0}^{\infty}\ln\left(x\right)\,\left(-{\rm e}^{-x} + 2{\rm e}^{-2x}\right)\,{\rm d}x
\\
& = &
\int_{0}^{\infty}\ln\left(x\right)\,{\rm e}^{-x}\,{\rm d}x
-
\int_{0}^{\infty}\ln\left(x \over 2\right)\,{\rm e}^{-x}\,{\rm d}x
=
\ln\left(2\right)\
\overbrace{\quad\int_{0}^{\infty}{\rm e}^{-x}\quad}^{1}
=
\ln\left(2\right)
\\[1cm]
&&\mbox{Or/and try this}
\\
I\left(\mu\right) & \equiv & \int_{0}^{1}{x^{\mu} - 1 \over \ln\left(x\right)}\,{\rm d}x\,,
\quad
I'\left(\mu\right) = \int_{0}^{1}{x^{\mu}\ln\left(x\right) \over \ln\left(x\right)}\,{\rm d}x
\\
I'\left(\mu\right)
& = &
\left.{x^{\mu + 1} \over \mu + 1}\right\vert_{0}^{1} = {1 \over \mu + 1}
\quad\Longrightarrow\quad
I\left(1\right) - I\left(0\right) = \int_{0}^{1}{{\rm d}\mu \over \mu + 1} = \ln\left(2\right)
\end{eqnarray*}
\begin{eqnarray*}
&&\\[1cm]
\int_{0}^{1}{\ln\left(x\right) \over x - 1}\,{\rm d}x
& = &
\int_{0}^{\infty}x{\rm e}^{-x}\,{1 \over 1 - {\rm e}^{-x}}\,{\rm d}x
=
\int_{0}^{\infty}x{\rm e}^{-x}\,\sum_{n = 0}^{\infty}{\rm e}^{-nx}\,{\rm d}x
=
\sum_{n = 0}^{\infty}
\int_{0}^{\infty}x\,{\rm e}^{-\left(n + 1\right)x}\,{\rm d}x
\\
& = &
\sum_{n = 0}^{\infty}{1 \over \left(n + 1\right)^{2}}\
\overbrace{\quad\int_{0}^{\infty}x\,{\rm e}^{-x}\,{\rm d}x\quad}^{1}
=
\sum_{n = 0}^{\infty}{1 \over \left(n + 1\right)^{2}}
\\
& = &
{\pi^{2} \over 6}\,,\quad\mbox{( Basel Problem )}
\end{eqnarray*}
http://en.wikipedia.org/wiki/Basel_problem.
Vea también: http://empslocal.ex.ac.uk/people/staff/rjchapma/etc/zeta2.pdf