= Solution
There are two different physical interpretations. The supplied logarithmic integral signals <free fall in a fixed singular isothermal sphere>, with a test particle moving through a frozen background. Write $C=\rho_0r_0^2$. By the <shell theorem>, the enclosed <mass> is $M(r)=4\pi Cr$ and the radial <equation of motion> is
$$
\ddot r=-\frac{4\pi GC}{r},\qquad
\frac{\dot r^2}{2}=4\pi GC\ln\frac{r_0}{r},\qquad \dot r<0.
$$
The second equality follows by multiplying by $\dot r$ and using the initial rest condition. The <free fall> trajectory is given implicitly by
$$
t(r)=\frac{r_0}{\sqrt{8\pi GC}}\int_{r/r_0}^1\frac{dx}{\sqrt{\ln(1/x)}}
=\frac{1}{\sqrt{8G\rho_0}}\operatorname{erf}\!\sqrt{\ln(r_0/r)}.
$$
Thus the intended fixed-profile answer is
$$
\boxed{t_{\rm ff,fixed}=\frac{1}{\sqrt{8G\rho_0}}.}
$$
\b[A collapsing material shell gives a different answer.] If the entire cloud undergoes pressureless <gravitational collapse> from rest, the edge follows the fixed enclosed <mass> $M_0=4\pi\rho_0r_0^3$, not the instantaneous fixed-profile value $4\pi Cr$. With no <shell crossing>, $\ddot r=-GM_0/r^2$, so
$$
t_{\rm ff,shell}=\int_0^{r_0}\frac{dr}{\sqrt{2GM_0(1/r-1/r_0)}}
=\frac{\pi}{2}\sqrt{\frac{r_0^3}{2GM_0}}
=\boxed{\sqrt{\frac{\pi}{32G\rho_0}}}.
$$
This is a concrete counterexample to treating the fixed-profile result as the exact collapse time of the whole nebula. The initial mean <density> inside $r_0$ is $3\rho_0$, consistent with the usual <free-fall time of a uniform sphere> expressed using that mean <density>. The logarithmic integral is valid for the explicitly frozen background approximation.
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