Differential equation $(u')^2(1+\alpha u^2)=1$ How to find $u$ such that (with constant $\alpha>0$ and $u'\ge 0$)
$$(u')^2(1+\alpha u^2)=1$$
By applying $u=\frac{\sinh(x)}{\sqrt{\alpha}}$, I found 
$$\cosh^2(x)=\sqrt{\alpha}.dx$$
and then $$d\left(\frac{2x+\sinh(2x)}{4}\right)=\sqrt{\alpha}.dx$$
not sure what to do after that. 
$2x+\sinh(2x)=(4\sqrt{\alpha}).t+C$, and then let 
$$f_\alpha(x)=\frac{2x+\sinh(x)}{4\sqrt{\alpha}}$$
Then $$u(t)=\sinh\left(f_\alpha^{-1}(t)\right)$$
But is there any way to have a more friendly expression for $u$, using usual functions ?

Take square root then substitute $\sqrt{\alpha} u = \sinh(x)$
