Extremely difficult log integral, real methods only

Question

$$\int_{0}^{1}\frac{x^2 + x\log(1-x)- \log(1-x) - x}{(1-x)x^2} dx$$

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.

Answer 1

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: $$ \int_0^1 \frac{1-\log(1-x)-1-x}{x^2} \mathrm dx = - \int_0^1 \frac{\log(1-x)+x}{x^2} \mathrm dx $$

Answer 2

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.