Trying to find a function from line length

Question

I was wondering if it was possible to find a function $f(x)$ whose line length is equal to its integral, thinking about it logically the function $f(x)=1$ satisfies these conditions as the line length is just equal to $x$ and so is the antiderivative, However I was wondering if there could be any more functions that satisfy this, that is to say: $$\int\sqrt{1+f'^2}\,dx=\int f\,dx$$ Which I think is fair to change to: $$\sqrt{1+f'^2}=f$$ $$1+f'^2(x)=f^2(x)$$ Any suggestions?

Answer 1

$$ \frac{df}{dx} = \pm \sqrt{f^2-1},\\ \frac{df}{\sqrt{f^2-1}} = \pm dx,\qquad\text{if $f\neq1$}\\ \frac{d(\cosh u)}{\sqrt{\cosh^2 u-1}} = \pm dx,\qquad f = \cosh u, \\ |u|=C\pm x,\\ f=\cosh(C\pm x) = \cosh(x+C). $$

So there are two solution $f_1=\cosh(x+C)$ and $f_2=1$. They can transform into each other at points where $f_1=f_2=1$, which for function $f_1$ happens only at one point $x=-C$, so a general solution is:

$$ f(x) = \begin{cases}\cosh(x-a), &x\le a,\\ 1,&a<x<b,\\ \cosh(x-b),& x\ge b\end{cases} $$

where $b\ge a$ are constants.