Showing $\int_{1}^{0}\frac{\ln(1-x)}{x}dx=\frac{\pi ^{2}}{6}$
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 $$\int\limits^{0}_{1}\frac{\ln(1-x)}{x}dx = 1 + \frac{1}{4} + \frac{1}{9}...$$
$$=\frac{\pi ^2}{6}$$
(I assume that what you mean is that you don't want to use the fact that $\sum 1/n^2=\pi^2/6$.)
Yes, there is a way of seeing this, and this is the basis of Mikael Passare's paper How to compute $\sum 1/n^2$ by solving triangles (free preprint on arXiv, or the published version on JSTOR for subscribers).
If you set $x=e^{-t}$, your integral becomes $$ \int_0^{\infty} -\ln(1-e^{-t}) \, dt , $$ or $$ \int_0^{\infty} -\ln(1-e^{-x}) \, dx $$ if we call the variable $x$ again. This is the area of a region in the first quadrant of the $xy$-plane, below the curve $y=-\ln(1-e^{-x})$ (or equivalently $e^{-x}+e^{-y}=1$). This is the region called $U_0$ i Passare's paper, and he shows that there are two other regions $U_1$ and $U_2$ with the same area, and then via a clever area-preserving change of variables that the combined area of $U_0$, $U_1$ and $U_2$ equals the area of a certain right triangle $T$ which is obviously $\pi^2/2$. Hence the area of $U_0$ (your integral) is $\pi^2/6$.
(See also this answer and this.)