As you can see, I asked this question here, and got a nice answer from the user eyeballfrog. Unfortunately, his answer, namely
$$
y = 2\tan^{-1}(x + C) – x + \frac{\pi}{2} + 2k\pi,
$$
doesn't match to the one in my textbook, i.e.
$$y=-2\tan^{-1}(1+\frac{2}{x+C})-x+2k\pi$$
I have tried it again with his tip multiple times this week but I didn't manage to get the textbook result.
Best Answer
Note that, upto a fixed multiple of $\pi$ (depending on where $x$ lies), the following holds. $$\tan^{-1}(u)+\tan^{-1}\left(1+\frac2{u-1}\right)=\tan^{-1}\left(\frac{u+1+\frac2{u-1}}{1-u\left(1+\frac2{u-1}\right)}\right)$$ The fraction inside nicely simplifies $$\frac{u+1+\frac2{u-1}}{1-u\left(1+\frac2{u-1}\right)}=\frac{u^2-1+2}{(u-1)-u(u+1)}=\frac{u^2+1}{-(u^2+1)}=-1$$ Therefore, $$2\tan^{-1}(x+C)+\frac{\pi}{2}=2m\pi-2\tan^{-1}\left(1+\frac2{x+C-1}\right)$$ This shows that the two solutions are equal for suitable relabelling of $C$ and $k$.
Hope this helps. :)