The value of ∫log(1−x)dx is
xlog(1−x)−12x−32x+C
(x−1)log(1−x)−12x−x+C
x2log(1−x)−12x+C
none of these
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.