I am solving the differential equation $$\left(y+y \sqrt{x^{2} y^{4}-1}\right) d x+2 x \ d y=0$$ and I have to use the substitution $u=xy^{2}$ to find the generic solution:
I have expressed $$y'=dy/dx = -\frac{y(1+ \sqrt{x^2y^4-1})}{2x}$$ I have $u'=y^{2}+2xy y'$, hence $u'=-y^{2}\sqrt{u^2-1}$. (I just substituted $y'$ from initial equation and substituted $u^{2}$ for $x^{2}y^{4}$.)
So $xu'=-u \sqrt{u^2-1}$. Then, $$\int \frac{du}{u \sqrt{(u^2-1)}}=\int \frac{-dx}{x} .$$
I solved the integral and got that the right-hand side is $-\ln|x|$, for the left side I used the substitution $u = \cosh t , d u = \sinh t\, dt$, which gives $\sqrt{u^2-1}=\sinh t$ and got the answer $\arctan(\sinh t)+c$, which is $\arctan({\sqrt{u^2-1}})+c$; but what do I have to do with it next?
Question: I don't understand the next step after i have $\arctan(\sqrt{u^2-1}) = -\ln|x|+c$
upd: Just saw this answer, which is only different because of sign under root, but I don't get how is their left integral $\int \frac{du}{u \sqrt{(u^2+1)}}$ is turned into $\ln|u|-\ln|1+\sqrt{u^2+1}|$?