Units within a log after integration

Question

Let $x(t)$ and $a$ be in unit meters, and $t$ unit time. If we look at the indefinite integral $$ \int\frac{\frac{dx}{dt}}{x(t) + a} dx $$ then we note that the units of the above expression are $\frac{meters}{seconds}$. If we evaluate the integral, we get $$ \frac{dx}{dt}\log(a + x(t)) $$ But the log has units meters in its argument, which, as I understand, transcendental functions' arguments should be dimensionless. Changing the indefinite integral into a definite one does not necessarily fix this, because, as I understand $$ \int_{x=x_0}^{x=x_1}\frac{\frac{dx}{dt}}{x(t) + a} dx = \frac{dx}{dt}\log(a + x(t))|^{x=x_1}_{x=x_0} = \frac{dx}{dt}(x=x_1)\log(a+x_1) - \frac{dx}{dt}(x=x_0)\log(a+x_0) $$ so what gives?