How to integrate$\int_0^1 \frac{\ln x}{x-1}dx$ without power series expansion

Here is to integrate without resorting to power series. Note \begin{align} I=\int_0^1\frac{\ln x}{1-x}& =\frac43\int_0^1 \frac{\ln x }{1-x}dx -\frac13\int_0^1 \overset{x\to x^2}{ \frac{\ln x }{1-x} } dx \\ &= \frac43\int_0^1 \frac{\ln x}{1-x^2}dx = \frac23\int_0^\infty \frac{\ln x}{1-x^2}dx=\frac23J(1) \end{align}

where $ J(\alpha) =-\frac 12 \int_0^\infty \frac{\ln (1-\alpha^2 + \alpha^2 x^2)}{x^2-1}dx $

$$ J'(\alpha) =-\int_0^\infty \frac{\alpha dx}{1-\alpha^2 + \alpha^2 x^2} = -\frac{\pi/2}{\sqrt{1-\alpha^2}}$$

Thus

$$ I= \frac23\int_0^1 J'(\alpha) d\alpha =-\frac{\pi}{3}\int_0^1 \frac{d\alpha}{\sqrt{1-\alpha^2}} =-\frac{\pi}{3} \sin^{-1}(1) = -\frac{\pi^2}{6}$$