Showing that $ \int_{0}^{1} \frac{x-1}{\ln(x)} \mathrm dx=\ln2 $ I would like to show that 
$$ \int_{0}^{1} \frac{x-1}{\ln(x)} \mathrm dx=\ln2 $$
What annoys me is that $ x-1 $ is the numerator so the geometric power series is useless.
Any idea?

$\displaystyle \int_{0}^{1}\frac{x-1}{\log{x}}\;{dx} = \int_{0}^{1}\int_{0}^{1}x^{t}\;{dt}\;{dx} =\int_{0}^{1}\int_{0}^{1}x^{t}\;{dx}\;{dt} = \int_{0}^{1}\frac{1}{1+t}\;{dt} = \log(2). $
