How do I integrate this metric? I have worked out a metric for a hypothetical pseudo-Riemann manifold and I'd like to explore its properties:
$$ds^2=-(at+c)^2dt^2+\left(\frac{\left(t_0^2+t_1^2\right)^2}{4t_0^4}\right)dx^2\tag 1$$
Where $ds$ is an infinitesimal distance (assume it's $0$ for all intervals), $t_1$ and $t_0$ are the start and end of the infinitesimal chunk of time, $dt$, at time, $t$. The change in the spatial coordinates comes from this schematic representation of the line elements:

$$dx=dx_0-\frac{1}{2}(dx_0-dx_1)\tag 2$$
I get the desired distance (that is, the sum total of all $dx$s) when I perform this operation numerically (as a series of time slices), but I'm having trouble making the conceptual leap to solving it using an integral (which, as I understand it, is the primary purpose of a metric). I suspect that I can replace $t_0^2+t_1^2$ with $t^2+(t-\Delta t)^2$. It also seems to me like $dt$ is conceptually equivalent to $\Delta t$, but I can't get any further.
 A: The classic concept of a differential has always been a troublesome idea in mathematics. Berkeley castigated them as "the ghosts of departed quantities", a humorous description that is none-the-less quite apt. As long as the quantity is still around, the ghost has no power. That is the issue with your "metric" - you are attempting to have the quantity and the ghost at the same time.
Mathematicians have invented several rigorous approaches to differentials, each with its own advantages and disadvantages. Without going into the weeds, perhaps the simplest is to consider any statement about differentials as being a differential equation on some curve. That is, there is some smooth curve $\gamma(u)$ on your manifold. $x$ and $t$ are functions on that manifold, so $x \equiv x\circ \gamma(u)$ and $t \equiv t\circ\gamma(u)$ are smooth functions on a real variable. $s$ is arclength along the curve from some initial point $\gamma(u_0)$, and as such $s$ is also a function of $u$. The differential equation becomes:
$$\left(\frac{ds}{du}\right)^2 = -(at + c)\left(\frac{dt}{du}\right)^2 + \left(\frac{\left(t_0^2+t_1^2\right)^2}{4t_0^4}\right)\left(\frac{dx}{du}\right)^2$$
(Because $s$ only appears in a derivative, the expression does not depend on $u_0$.) Now $t$ is still this function of the location in space, and therefore well defined for any $u$, but what are $t_0$ and $t_1$? How do they depend on $u$?
We can take one of them to be $t$ (it doesn't matter which). But the other is supposed to differ from it by "$\Delta t$". What $\Delta t$? There is no such parameter. Differential equations occur at single points, not between two points. They are often obtained as an infinitesimalized version of a delta equation between two points, but that infinitesimalization always reduces to the behavior at a single point, not two. Only the lowest order terms in the deltas survive it to become differentials. Higher order terms all go to $0$. In the case of your $t_0$ and $t_1$, if we set $t_0 = t$ and $t_1 = t + \Delta t$, then the lowest order term in $\frac{\left(t_0^2+t_1^2\right)^2}{4t_0^4}$ is just $1$. This naive guess would make your metric
$$ds^2 = -(at + c)dt^2 + dx^2$$
But I don't know how you arrived at that expression, and a more careful analysis of your situation may provide a different answer. But if done correctly it will not involve $t_0$ and $t_1$.
