You have to specify your domain.
For log(x-1) you need x>1
For log(1-x) you need x<1
May times you will find it written that
∫ dx/x = log |x| + C
just for this reason.
Integrate x dx/(1-x). I have proceeded thus-
Int xdx/(1-x)=int -(x-1+1)/(x-1)
=-Int[1+ 1/(x-1)]dx
=-Int dx-Int dx/(x-1)
=-x-log(x-1). On differentiating, we get original expression-
d/dx[-x-log(x-1)]=-1-1/(x-1)=-x/(x-1)=x/(1-x).
However, the answer in the book is
-x-log(1-x)and differentiating this also we get same expression-
d/dx[-x-log(1-x)]=-1+1/(1-x)=x/(1-x).
There are no constants of integration in this example and log(1-x)is not=log(x-1), then where is the anomaly?
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