Evaluate $ \int_0^1\frac{\log(x)}{x-1}dx $

Question

Can I evaluate this integral:

$$ \int_0^1\frac{\log(x)}{x-1}dx $$

without knowing the value of $\zeta(2)$? In particular can I use methods of complex analysis?

Answer 1

$\begin{align} \int_0^1 \frac{\ln x}{x-1}\,dx&=\Big[\ln(1-x)\ln x\Big]_0^1-\int_0^1 \frac{\ln(1-x)}{x}\,dx\\ &=-\int_0^1 \frac{\ln(1-x)}{x}\,dx\end{align}$

and read my answer, https://math.stackexchange.com/a/2632547/186817

Answer 2

$$-\int^{1}_{0}\frac{\ln x}{1-x}dx=-\int^{1}_{0}\ln x\sum^{\infty}_{n=0}x^ndx$$

$$=-\sum^{\infty}_{n=0}\int^{1}_{0}x^n\ln(x)dx$$

By parts method

$$ \sum^{\infty}_{n=0}\frac{1}{(n+1)^2}=\frac{\pi^2}{6}=\zeta(2)$$