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 ?
 A: Hint 
You could notice that the equation is separable since 
$$\left(\frac{du}{dx}\right)^2 (1+\alpha u^2)=1$$ 
can write $$\frac{dx}{du}=\pm \sqrt {1+\alpha u^2}$$ Now, use user121049's suggestion for an appropriate change of variable for the integration of the rhs.
Added later to this answer
The final solution to this problem is $$x=\frac{1}{2} u \sqrt{\alpha  u^2+1}+\frac{\sinh ^{-1}\left(\sqrt{\alpha } u\right)}{2    \sqrt{\alpha }}$$ Concerning the asymptotic behavior, the OP showed by himsel that, for large values of $x$ can be derived the approximation $$ u \simeq \frac{\sqrt{2x} }{\sqrt[4]{\alpha }}$$ It is possible to obtain a better approximation developing the rhs as a Taylor series built at $x=\infty$. This leads to $$x \simeq \frac{\sqrt{\alpha } u^2}{2}+\frac{\log (4 \alpha )+2 \log (u)+1}{4 \sqrt{\alpha }}$$ Reversing this last expression leads to $$u \simeq \frac{\sqrt{W\left(\frac{1}{2} e^{4 \sqrt{\alpha } x-1}\right)}}{\sqrt{2 \alpha}
  }$$ which is much better.
A: Take square root then substitute $\sqrt{\alpha} u = \sinh(x)$
