Integral Equation (no ideia) involving integral of curvature. $$ \int_0^x \frac1{c-f(t)}dt = \frac {f'(x)}{\sqrt{(f'(x))^2+1}}-1$$
either estimation, or general solution.
what ever c that make's it easier, of course not including zero.
Thank you.
 A: I'd like to explain here in more detail what I wrote in a comment.
We have to solve the equation
$$ \int_0^x \frac1{c-f(t)}dt = \frac {f'(x)}{\sqrt{(f'(x))^2+1}}-1\tag{1}$$
Taking the derivative with respect to $x$ gives
$$\frac{1}{c-f(x)}=\frac{d}{dx}\left( \frac{f'(x)}{\sqrt{f'(x)^2+1}}-1\right)= \frac{f''(x)}{\left(f'(x)^2+1\right)^{3/2}}\tag{2}$$
This is a second order ODE.
Now multiplying $(2)$ with $f'(x)$ can be written as
$$\frac{f'(x)}{c-f(x)}=-\frac{d}{dx}\log (c-f(x))=\frac{f'(x) f''(x)}{\left(f'(x)^2+1\right)^{3/2}}=\frac{d }{dx}\frac{f'(x)}{\sqrt{f'(x)^2+1}}$$
whence we find that
$$-\log (c-f(x))+\frac{f'(x)}{\sqrt{f'(x)^2+1}}=c\tag{3}$$
with a constant $c$. $(3)$ is a first integral of $(2)$ and represents a separable ODE of the first order.
Solving $(3)$ for $f'(x)$  gives
$$f'(x) = \frac{df}{dx} = \pm \frac{a-\log (c-f(x))}{\sqrt{1-(a-\log (c-f(x)))^2}}\tag{4}$$
this can be immediately integrated to give the inverse function of $f(x)$
$$x(f) = \pm \int\,df \sqrt{\frac{1}{(a-\log (c-f))^2}-1}\tag{5}$$
Discussion
The numerical solutions of $(2)$ for $c=1$ with $f(0)=2$ and high positive values of $f'(0)$, say $=50$, suggests that $f(x)$ describes an ellipse.
