Why does $\int\limits_0^1 {\dfrac{{x - 1}}{{\ln x}}} \;\text{d}x=\ln2$?

Question

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

Answer 1

$$t>-1:$$

$$\begin{aligned}f(t)=\int_0^1 \frac{x^t-1}{\ln x}\,dx\;\Rightarrow\; f'(t) &=\int_0^1 x^t\,dx =\frac{1}{t+1}\end{aligned}$$

$$f(1)=\int_0^1 \frac{dt}{t+1}=\ln 2$$