TITLE: Why does $\int\limits_0^1 {\dfrac{{x - 1}}{{\ln x}}} \;\text{d}x=\ln2$?
QUESTION [6 upvotes]: I have found that
$$\int\limits_0^1 {\dfrac{{x - 1}}{{\ln x}}} \;\text{d}x=\ln2$$
but I can't prove it.
Any hint?
Thank you in advance

REPLY [8 votes]: $$t>-1:$$
$$\begin{aligned}f(t)=\int_0^1 \frac{x^t-1}{\ln x}\,dx\;\Rightarrow\; f'(t) &=\int_0^1 x^t\,dx =\frac{1}{t+1}\end{aligned}$$
$$f(1)=\int_0^1 \frac{dt}{t+1}=\ln 2$$