Is it true that the definite integral of $\dfrac1{v(x)}$ with respect to $x$ equals the time, where $v(x)$ is velocity as a function of distance?
I intuitively think that’s true because if you take a small distance $\mathrm dx$ where the velocity remains constant, the time spent equals $\dfrac{\mathrm dx}{v(x)}$. Then, the total time would be the integral of that from the value of $x$ to another.
We know that $$ \left\{ \matrix{ x = f(t) \hfill \cr dx = {{df(t)} \over {dt}}dt = v(t)dt \hfill \cr} \right. $$
Now, if $f(t)$ is invertible, or for each time interval in which it is , we can write $$ \left\{ \matrix{ t = f^{\,\left( { - 1} \right)} (x) \hfill \cr dt = {{df^{\,\left( { - 1} \right)} (x)} \over {dx}}dx = \hfill \cr {{df^{\,\left( { - 1} \right)} (x)} \over {dx}}{{df(t)} \over {dt}}dt = {{df^{\,\left( { - 1} \right)} (x)} \over {dx}}v(f^{\,\left( { - 1} \right)} (x))dt \hfill \cr} \right. $$ which means that it must be $$ \left\{ \matrix{ {{df^{\,\left( { - 1} \right)} (x)} \over {dx}}{{df(t)} \over {dt}} = 1 \hfill \cr {{df^{\,\left( { - 1} \right)} (x)} \over {dx}}v(f^{\,\left( { - 1} \right)} (x)) = 1 \hfill \cr} \right. $$
You can conclude by yourself that your intuition is true.