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).