<s>[INST] 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. 
 [/INST] You can write the numerator as :
$$
x^2 +x\log(1-x) -\log(1-x)-x = (1 - \log(1-x))(1-x) +(x-1)(x+1)
$$
so that it remains to evaluate the integral:
$$
\int_0^1 \frac{1-\log(1-x)-1-x}{x^2} \mathrm dx = - \int_0^1 \frac{\log(1-x)+x}{x^2} \mathrm dx
$$
</s>