Schwarzschild metric in Isotropic coordinates As one wants to jump to Isotropic coordinates in order to write the Schwarzschild metric in terms of them, one does this coordinate transformation:
$$r=r'\left(1+\frac{M}{2r'}\right)^2$$
So we start with the very well-known form:
$$ds^2 = -\left(1-\frac{2m}{r}\right)dt^2 + \left(1-\frac{2m}{r}\right)^{-1}dr^2 +r^2(d\theta^2 +\sin^2\theta d\phi^2) $$
And arrive at 
$$ ds^2 = -\left(\frac{1-M/2r'}{1+M/2r'}\right)^2dt^2 +(1+M/2r')^4[dr'^2 +r'^2( d\theta^2 +\sin^2\theta d\phi^2)]$$
My question is: Where did this coordinate transformation come from? 
 A: The aim of the isotropic coordinates is to write the metric in the form where the spacelike slices are as close as possible to Euclidean. That is, we try to write the metric in the form:
$$ ds^2 = -A^2(r)dt^2 + B^2(r)d\Sigma^2 $$
where $d\Sigma^2$ is the Euclidean metric:
$$ d\Sigma^2 = dr^2 + r^2(d\theta^2 + \sin^2\theta d\phi^2) $$
So let's use the substitution $r\rightarrow r'$ and write down our metric:
$$ ds^2 = -\left(1-\frac{2M}{r'}\right)dt^2 + B^2(r')\left(dr'^2 + r'^2(d\theta^2 + \sin^2\theta d\phi^2)\right) $$
If we compare this with the Schwarzschild metric:
$$ ds^2 = -\left(1-\frac{2M}{r}\right)dt^2 + \frac{dr^2}{1 - 2M/r} + r^2(d\theta^2 + \sin^2\theta d\phi^2) $$
Then for the angular parts to be equal we must have:
$$ B^2(r')r'^2 = r^2 $$
And for the radial parts to be equal we must have:
$$ B^2(r')dr'^2 = \frac{dr^2}{1 - 2M/r} $$
Divide the second equation by the first to eliminate $B$ and we end up with:
$$ \frac{dr'^2}{r'^2} = \frac{dr^2}{r^2 - 2Mr} $$
And then just take the square root and integrate and we get the substitution you describe:
$$ r = r'\left(1 + \frac{M}{2r'}\right)^2 $$
