The value of ∫log⁡(1−x)dx is

Integrating by parts taking log (1−x) as the first function, the given integral can be written as xlog⁡(1−x)+12∫x1−xdx=xlog⁡(1−x)+∫t21−tdt−xlog⁡(1−x)+∫−t−1+11−tdt=(x−1)log⁡(1−x)-x2−x+C.