Verify solution to ODE I am given the ODE
$$\left(f''(x)+\frac{f'(x)}{x} \right) \left(1+f'(x)^2 \right) = f'(x)^2f''(x)$$ 
and I already know that the solution to this ODE is given by
$$f(x)= c  \cdot arcosh \left( \frac{r}{c} \right) + d$$
where $|c|<r$ and $d \in \mathbb{R}.$
The problem is I want to show that this is an actual solution by direct integration (so I want to derive it) and not just verify it by plugging it in. Does anybody know how this can be done?
After rearraging as Daniel Fischer proposed, I end up with
$$f''(r) = -\frac{(f'(r)^3+f'(r))}{r}$$

$$
\left(f'' +\frac{f'}{x}\right)(1+f'^2)=f'^2f''
$$
multiply by $x$
$$
(xf''+f')(1+f'^2) = \left(\dfrac{d}{dx}xf'\right)(1+f'^2) = xf'^2f''
$$
then we have
$$
\frac{1}{xf'}\left(\dfrac{d}{dx}xf'\right) = \frac{f'f''}{(1+f'^2)}
$$
thus
$$
\dfrac{d}{dx}\ln(xf') = \frac{1}{2}\dfrac{d}{dx}\ln(1+f'^2)
$$
which is
$$
\ln (xf') = \frac{1}{2}\ln(1+f'^2) + C
$$
or
$$
x^2f'^2 = C_1(1+f'^2)
$$
or
$$
(x^2-C_1)f'^2 = C_1\implies f' = \sqrt{\frac{C_1}{x^2-C_1}}
$$
or
$$
\dfrac{df}{dx} = \frac{1}{\sqrt{\left(\frac{x}{\sqrt{C_1}}\right)^2-1}}
$$
you should see what the solution of the last equation is.
