I am dealing with this problem: Asymptotic Curves and Lines of Curvature of Helicoid. It's the question 3.3.2 from Do Carmo's book Differential Geometry of Curves and Surfaces.
I done the calculus finding:
$$e=g=F=0\\ E=v^2+c^2\\ G=1\\ f=\dfrac{c}{\sqrt{v^2+c^2}}$$
So, putting the values into the equation for curvature lines:
$$\begin{vmatrix}(v')^2& -u'v'& (u')^2\\ E& F& G\\ e& f& g\end{vmatrix}= 0,$$
I've got:
$$(u')^2(v^2+c^2)=(v')^2$$
$$(u')\sqrt{v^2+c^2}=\pm v'$$
$$\log((u')\sqrt{v^2+c^2}\pm v')=ctc.$$
Maybe it is obvious, but I am not understanding how go from here to the book answer: $\log(v+\sqrt{v^2+c^2})\pm u=ctc.$ Or maybe my calculus are wrong...
Many thanks in advance for some clue.