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The value of log(1x)dx is 

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a
xlog⁡(1−x)−12x−32x+C
b
(x−1)log⁡(1−x)−12x−x+C
c
x2log⁡(1−x)−12x+C
d
none of these

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detailed solution

Correct option is B

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.


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