separable differential equation

I can't seem to get the solution in my text book, the problem is:

$\displaystyle xy' + y = y^{2}$

I began by reorganizing to:

$\displaystyle \frac{1}{y(y-1)}dy = \frac{1}{x}dx$

expanding the y term and integrating:

$\displaystyle \int{\frac{1}{y}dy} - \int{\frac{1}{y-1}dy} = \int{\frac{1}{x}dx}$

giving the following:

$\displaystyle ln|y| - ln|y-1| = ln|x| + K$

where K is the combined constant of integration from both sides

now when I take the exponential, I think I have to raise e to the entire

right side and that I can't do it term by term which simplifies to a division

using log properties and then canceling out the e and ln

$\displaystyle |\frac{y}{y-1}| = |x|e^{K}$

now I was thinking I could drop all the absolute values if I throw a +\-

in front of the remaining exponential, so we'll say $\displaystyle C=\pm e^{K}$

$\displaystyle \frac{y}{y-1} = Cx$

solving for y yields:

$\displaystyle y = \frac{-Cx}{1-Cx}$

however my textbook gives:

$\displaystyle y = \frac{1}{1-Cx}$

Please let me know if you can see my mistake, thanks