Showing $\int_{1}^{0}\frac{\ln(1-x)}{x}dx=\frac{\pi ^{2}}{6}$ Is there way to show $$\int_{1}^{0}\frac{\ln(1-x)}{x}dx=\frac{\pi ^{2}}{6}$$ without using the Riemann zeta function?

Note: As pointed out by Hans, this answer uses the fact that $\zeta(2) = \dfrac{\pi ^2}{6}$ and therefore isn't what the OP asked for.
$$\frac{\ln(1+x)}{x} = 1 - \frac{x}{2} + \frac{x^2}{3}...$$
$$\int\frac{\ln(1+x)}{x}dx = x - \frac{x^2}{4} + \frac{x^3}{9}...+C$$
$$\int\limits^{0}_{1}\frac{\ln(1+x)}{x}dx = -1 + \frac{1}{4} - \frac{1}{9}...$$

After question edit:
We have $$\frac{\ln(1-x)}{x} = - 1 + \frac{x}{2} - \frac{x^2}{3}...$$
$$\int\frac{\ln(1-x)}{x}dx = -x + \frac{x^2}{4} - \frac{x^3}{9}...+C$$
Therefore $$\int\limits^{0}_{1}\frac{\ln(1-x)}{x}dx = 1 + \frac{1}{4} + \frac{1}{9}...$$
$$=\frac{\pi ^2}{6}$$
