How do I solve $\int \frac{1}{v(1+v^2)}dv$? I'm trying to solve $y'=\frac{2xy}{x^2-y^2}$ and I ended up with the integral $\int \frac{1}{v(1+v^2)}dv$ as part of the solution.

I got a big ugly solution using integration by parts, but I'm hoping there's a simpler way.
Now use partial decomposition:

$$\int\frac{1}{v(1+v^2)}dv = \int[\frac{1}{v} - \frac{v}{1+v^2}]dv$$

$$=\ln|v|-\frac{1}{2}\ln(1+v^2)+C$$
$$\log x -\log(1+x^2)/2$$

does it, I think.