Stuck on integrating $\int x/(1-x)dx$

Indefinite integrals always have an extra constant term. The general answer is $-x - \ln|1-x| + C$. If you add a constant term to both your answer and the one you said it is supposed to be, you will find that they are equivalent.

$$ \displaylines{ \int {\frac{x}{{1 - x}}dx} = \int {\frac{{x + 1 - 1}}{{1 - x}}dx} \cr = \int {\left( { - 1 + \frac{1}{{1 - x}}} \right)dx} \cr = - x - \ln \left| {1 - x} \right| + c \cr} $$