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?

This is a classic example of differentiating inside the integral sign.
In particular, let $$J(\alpha)=\int_0^1\frac{x^\alpha-1}{\log(x)}\;dx$$. Then one has that $$\frac{\partial}{\partial\alpha}J(\alpha)=\int_0^1\frac{\partial}{\partial\alpha}\frac{x^\alpha-1}{\log(x)}\;dx=\int_0^1x^\alpha\;dx=\frac{1}{\alpha+1}$$ and so we know that $\displaystyle J(\alpha)=\log(\alpha+1)+C$. Noting that $J(0)=0$ tells us that $C=0$ and so $J(\alpha)=\log(\alpha+1)$.
