Derivation of Proper Time of Fall in Schwarzschild Metric Some time ago, in this question (Proper Time of Fall in Schwarschild Metric), I asked how to find the proper time of fall of an observer in the Schwarzschild metric because I had found many different ways of deriving it and they did not seem to mean the same thing.
There @Yukterez said the proper acceleration of an infalling observer is:
$$\frac{d^{2}r}{d\tau^{2}} = -\frac{GM}{r^{2}}$$
What I understand now. You can find this by solving the Geodesic Equation for radial fall. He said I could solve this differential equation by integrating from $\frac{d^{2}r}{d\tau^{2}}$ to $\frac{dr}{d\tau}$ and then the reciprocal, but I do not understand how this leads to the expression he gave afterwards. Which is:
$$\rm \tau=\frac{r_0^{3/2}arctan\left(\sqrt{r_0/r-1}\right)}{c \ \sqrt{r_s}}+\frac{r \ r_0\sqrt{r_s/r-r_s/r_0}}{c \ r_s}$$
After this, @samuel-adrian-antz helped me by showing a method of solving that differential equation that I was able to fully understand.
I thank them both for helping me, but here I'd like to ask how to reach Yukterez's answer, because I see many texts giving it as answer. But also, why there's so many different ways of stating the proper time of fall? And what all these aproaches mean? Are each valid only for some cases?
My biggest doubt might be why direct integration do not seem to work  here propely when I try it. I know this differential equation is for $\frac{d^{2}r}{d\tau^{2}}$, and I need $\frac{d\tau}{dr}$, but then how to reach it since I do not know how $r$ depends on $\tau$?
 A: Lets walk through the derivation, starting from
$$\frac{d^{2}r}{d\tau^{2}} = -\frac{GM}{r^{2}}.$$
We can multiply both side with $\frac{dr}{d\tau}$,
$$\frac{d^{2}r}{d\tau^{2}}\frac{dr}{d\tau} = -\frac{GM}{r^{2}}\frac{dr}{d\tau}.$$
We now apply the chain rule to rewrite each side as a total derivative w.r.t. $\tau$,
$$\frac{1}{2}\frac{d}{d\tau} \left(\frac{dr}{d\tau}\right)^2 = \frac{d}{d\tau} \frac{GM}{r}.$$
Since both sides are total derivatives we can integrate once w.r.t. $\tau$ to get
$$\frac{1}{2} \left(\frac{dr}{d\tau}\right)^2 =  \frac{GM}{r}-\frac{GM}{r_0},$$
where have used the condition $\frac{dr}{d\tau}=0$ at $r=r_0$.
With some rearranging we get,
$$ \frac{dr}{d\tau} = -\sqrt{ \frac{2GM(r-r_0)}{r_0r}}.$$
We can now apply the inverse function theorem, to get
$$ \frac{d\tau}{dr} = -\sqrt{ \frac{r_0r}{2GM(r-r_0)}}.$$
We now have a first order ODE for $\tau$ as a function of $r$, with no explicit dependence on $\tau$ on the right hand side, so this can be solved by direct integration to get the answer.
