Extremely difficult log integral, real methods only $$\int_{0}^{1}\frac{x^2 + x\log(1-x)- \log(1-x) - x}{(1-x)x^2} dx$$
I tried this:
$$M_1 = \int_{0}^{1} \frac{1}{1-x} \cdot \left(\frac{x^2 + x\log(1-x) - \log(1-x) - x)}{x^2}\right) dx$$
$$M_1 = \int_{0}^{1} \frac{x^2 + x\log(1-x) - \log(1-x) - x}{(1-x)x^2} dx$$
$$M_1 = \int_{0}^{1} \frac{x\left(x + \log(1-x) - 1\right) - \log(1-x)}{(1-x)x^2} dx$$
$$M_1 = \int_{0}^{1} \frac{x + \log(1-x) - 1}{x(1-x)}dx - \int_{0}^{1} \frac{\log(1-x)}{(1-x)x^2} dx$$
but we cannot seperate as the integrals become divergent. 

Using Paul_I's answer and the classic expansion 
$$\log\left(1-x\right)=-\sum_{n=1}^\infty \frac{x^n}{n}, \quad -1<x<1,$$
we get
$$-\int_0^1\frac{\log(1-x)+x}{x^2}\mathrm dx =\sum_{n=2}^\infty\int_0^1 \frac{x^{n-2}}{n}\mathrm dx = \sum_{n=1}^\infty\frac{1}{n(n+1)}=1$$
in accordance with Mathematica.
