= Solution
The metric is nonsingular on $r>0$; $r=0$ is a coordinate boundary at which the displayed coefficients diverge. The nonzero <Christoffel symbols> are
$$
\Gamma^t_{tr}=\Gamma^t_{rt}=-1/r,\qquad \Gamma^r_{tt}=-1/r,\qquad \Gamma^r_{rr}=-1/r.
$$
Writing dots for derivatives with respect to <proper time>, the <geodesic equations> are
$$
\boxed{\ddot t-2\dot t\dot r/r=0,\qquad \ddot r-(\dot t^2+\dot r^2)/r=0.}
$$
The first gives $\dot t=kr^2$ for a constant $k$, chosen positive for a future-directed particle. Normalization of a timelike four-velocity gives $(-\dot t^2+\dot r^2)/r^2=-1$, so $\dot r^2=k^2r^4-r^2$. Dividing by $\dot t^2$ gives
$$
\boxed{(dr/dt)^2=1-1/(k^2r^2).}
$$
Integrating on each side of the turning point yields $\sqrt{r^2-k^{-2}}=|t-t_0|$, hence $r^2-(t-t_0)^2=k^{-2}$. Choose $t_0=0$ and let $\tau=0$ at the minimum radius. Then
$$
\boxed{kr=\sec\tau,\qquad kt=\tan\tau,\qquad -\pi/2<\tau<\pi/2.}
$$
Differentiating verifies $\dot t=\sec^2\tau/k=kr^2$, the normalization and the radial <Geodesic equation>. <proper time> therefore remains finite as the coordinate radius and time tend to infinity at the endpoints of this chart.
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