Removing a Coordinate Singularity of a 2D metric While trying to find the null geodesics of the metric
$$
ds^2 = (r^2 - 1)dt^2 - (drdt + dtdr)
$$
gives $$\frac{dt}{dr} = \frac{2}{r^2-1}$$ which is singular at $r=1$. However, we know that this is a coordinate singularity because the curvature (Ricci) scalar is independent of $r$, $$\mathcal{R} = 2$$
So, how can I get the correct null geodesics? Or what is the coordinate transformation needed here?
 A: Your equation is just
$$\frac{dr}{dt} = \frac{1}{2}(r^2-1) \implies \frac{dr}{r^2-1} = \frac{dt}{2}$$
This expression can readily be integrated to yield a non-singular $r(t)$.  To more specifically address your concern, if you set $r=1$ then the null geodesics are simply given by $r(t)=1$, since $\mathrm dr/\mathrm dt=0$.
A: I am not sure if this answer your question ?
the metric from your line element is:
$$G= \left[ \begin {array}{cc} {r}^{2}-1&-1\\ -1&0
\end {array} \right]
$$
the determinate of the metric is  $~\text{det}(G)=-1~$ hence the metric is not singular at the entire space.

we are looking for the transformation matrix $~\mathbf{T}~$ that fulfilled  this matrix equation
\begin{align*}
&\mathbf{T^T}\,\mathbf G\,\mathbf{T}=\begin{bmatrix}
                                       1 & 0 \\
                                       0 & -1 \\
                                     \end{bmatrix}\quad (1)\quad
\text{4 equations for the unknowns $~T_i~,~i=1..4~$ }  \\                                   
&\text{solution}\\
&\mathbf{T}= \left[ \begin {array}{cc} 1&1\\ \frac 12\,{r}^{2}-1&\frac 12
\,{r}^{2}\end {array} \right]\\\\
&\text{hence with}\\
&\begin{bmatrix}
   dt \\
   dr \\
 \end{bmatrix}=\mathbf{T}
 \begin{bmatrix}
   d\tau \\
   du \\
 \end{bmatrix}\quad\Rightarrow\quad
\left( {r}^{2}-1 \right) {{\it dt}}^{2}-2\,{\it dt}\,{\it dr}\mapsto
d\tau^2-du^2                            
\end{align*}
