ok you win, this was less algebra, but I didn't see it...so I guess... idunno..
That would have given, $du/dt = g'(t)$, and so $dt = du / g'(t)$.
From there $$
\int{\frac{\partial}{\partial t} g(t)\frac1{g(t)}}\, \mathrm{d}t =
\int{g'(t)\frac1u}\, \frac{\mathrm{d}u}{g'(t)} = \int{\frac1u}\, \mathrm{d}u = \ln|g(t)| + C$$