Indefinite integral $\int\ln(1−x)dx$ by parts

Question

I have an exercise that explicitly asks for the indefinite integral of the following equation using integral by parts. $$ \int\ln(1-x)\,dx $$ I tried to define: $ u = \ln(1-x) $ so, $$ \frac{\,du}{\,dx} = \frac{-1}{1-x}\\ \,du = \frac{-\,dx}{1-x} $$
and, $ \,dv = \,dx $ resulting in $ v = x $ and finally applying: $ \int{u\,dv} = uv - \int{v\,du} $ $$ \begin{align} \int\ln(1-x)\,dx&= x \ln(1-x) -\int x \, \frac{-\,dx}{1-x}\\ &=x\ln(1-x) - \int \frac{-x}{1-x} \,dx\\ &=x\ln(1-x) - \int \frac{1-x-1}{1-x} \,dx\\ &=x\ln(1-x) - \int \left(\frac{1-x}{1-x} - \frac{1}{1-x} \right)\,dx\\ &=x\ln(1-x) - \int \left(1 - \frac{1}{1-x} \right)\,dx\\ &=x\ln(1-x) - \left(\int{1\,dx} - \int{\frac{1}{1-x}}\,dx\right)\\ &=x\ln(1-x) - x -\left(- \int{\frac{1}{1-x}}\,dx\right)\\ &=x\ln(1-x) - x + \int{\frac{1}{1-x}}\,dx\\ &=x\ln(1-x) - x + \ln(1-x) + c\\ &=(x + 1)\ln(1-x) - x + c \end{align} $$
that does not match the answer in the book: $(x - 1)\ln(1-x) - x + c$
I have found this other solution that I believe is not valid considering that the exercise asks to utilize integral by parts.

Answer 1

Your steps all ALL correct except the last second step as @J.G. said.

On the other hand, using integration by parts wisely, we get $$ \begin{aligned} \int \ln (1-x) d x & =-\int \ln (1-x) d(1-x) \\ & =(1-x) \ln (1-x)+\int(1-x) \frac{1}{1-x} d x \\ & =(1-x) \ln (1-x)+x+C \end{aligned} $$

Answer 2

The mistake was to consider that $\int{\frac{\,dx}{1-x}}$ is simply $\ln(1-x)$ when the correct path of continuing is using the substitution method resulting in $-\ln(1 - x) + c$:
$ t = x - 1 $
$ \frac{\,dt}{\,dx} = -1$
$ \,dt = -\,dx$
$ \,dx = -\,dt$ $$ \int{\frac{-\,dt}{t}} = -1\int{\frac{\,dt}{t}} = -1\ln(1-x) + c $$