Solve this equation for the function $y(x)$:
$y' = \alpha \left(\int\sqrt{1 + y'^2} dx \right)^2$
Of course this must first be solved for $y'$ and then integrated to get $y$.
The following is not directly relevant to the problem, but gives the context in which it came up: Consider a road of $n$ lanes where the velocities of cars on successive lanes form an arithmetic progression. Now there is a turn of radius taken equal across all lanes. Each lane is banked so as to have no friction (then it must have the angle $\theta$ where $\frac{v^2}{r}=g\tan\theta$). Now take $n$ to infinity, keeping each lane of equal width and holding the minimum and maximum velocities (on the left and right sides of the road) fixed. Now $\tan\theta=\frac{dy}{dx}$ is proportional to $v^2$, and $v$ is proportional to $\int\sqrt{1 + y'^2} dx$.