In the case of a diagonal metric, \begin{align} \mathrm{d}s^2=g_{\mu\nu}\mathrm{d}{x}^\mu\mathrm{d}{x}^\nu, \end{align} it is relatively straightforward to find the Christoffel symbols by comparing the Euler-Lagrange equation \begin{align} \frac{\mathrm{d}}{\mathrm{d}\tau}\left(\frac{\partial L}{\partial \dot{x}^\mu}\right)-\frac{\partial L}{\partial x^\mu}=0, \end{align} where $L=\frac{1}{2}g_{\mu\nu}\dot{x}^\mu\dot{x}^\nu$ and $\dot x^\mu=\mathrm{d}x^\mu/\mathrm{d}\tau$, to the geodesic equation \begin{align} \ddot{x}^\mu+\Gamma_{\rho\sigma}^\mu\dot x^\rho\dot x^\sigma=0. \end{align}
However, this becomes less straightforward for a metric with non-diagonal terms. Additional cross terms in the line element will make not one but two second order derivative terms appear in the Euler-Lagrange equation, making a direct comparison to the geodesic equation less insightful.
Consider for illustrative purposes a 2 dimensional metric \begin{align} \mathrm{d}s^2=f\mathrm{d}t^2+g\mathrm{d}t\mathrm{d}r+h\mathrm{d}r^2, \end{align} with arbitrary functions $f=f(t,r),g=g(t,r),h=h(t,r)$.
In this case the $\mu=t$ and $\mu=r$ components give for the Euler-Lagrange equations respectively \begin{align} 2f\ddot t+g\ddot r+2\left(\frac{\partial f}{\partial t}\dot t+\frac{\partial f}{\partial r}\dot r\right)\dot t+\left(\frac{\partial g}{\partial t}\dot t+\frac{\partial g}{\partial r}\dot r\right)\dot r-\frac{\partial f}{\partial t}\dot t^2-\frac{\partial g}{\partial t}\dot t\dot r-\frac{\partial h}{\partial t}\dot r^2=0\\ 2h\ddot r+g\ddot t+2\left(\frac{\partial h}{\partial t}\dot t+\frac{\partial h}{\partial r}\dot r\right)\dot r+\left(\frac{\partial g}{\partial t}\dot t+\frac{\partial g}{\partial r}\dot r\right)\dot t-\frac{\partial f}{\partial r}\dot t^2-\frac{\partial g}{\partial r}\dot t\dot r-\frac{\partial h}{\partial r}\dot r^2=0. \end{align}
Now the additional $g\ddot r$ in the first and $g\ddot t$ in the second equation forbid a direct comparison to the geodesic equation and subsequently finding the Christoffel symbols.
How do we in general find the Christoffel symbols for a metric with non diagonal terms this way? Is it as simple as substituting one Euler-Lagrange equation in the other to eliminate either of the second order derivative terms?