Partial Integration - Where did I go wrong? For a Homework, I need $\int \frac{x}{(x-1)^2} dx$ as an intermediate result. Using partial integration, I derive $x$ and integrate $\frac{1}{(x-1)^2}$, getting: $$ \frac{-x}{x-1} + \int \frac{1}{x-1} dx = \ln(x-1)+\frac{x}{x-1} $$
WolframAlpha tells me this is wrong (it gives me $\frac{1}{1-x}$ where I have $\frac{x}{x-1}$). If I and WA disagree the error is usually somewhere on my side. Unfortunately WA uses partial fractions there instead of partial integration, so I'm not sure which step I screwed up. Supposedly $\int f'g dx =  fg - \int fg' dx$ right?
(I leave the constant +C out because it's not relevant for the problem I need this for).

the  result   is  :   $\ln|x-1| - \frac x{x-1}  +  C$   where  $C$  is  some  constant.
If  $C=1$   you  get   :  $\ln |x-1| +  \frac 1{1-x}$ 
The  finaly  result  can  be  expressed :
$$F(x) = \ln |x-1| +  \frac 1{1-x} +  \lambda$$   where $\lambda$  is  some constant.
Precesely :
 $$F(x) = \ln (x-1) +  \frac 1{1-x} +  \lambda$$ on  the  intervale $]1,+\infty[$
and:
$$F(x) = \ln (1-x) +  \frac 1{1-x} +  \lambda$$   on  the  intervale: $]-\infty ,1[$
