Let $f(t) = \int\_0^t \ln{(s^2+t^2)} ds$, how can I find the derivative $\frac{df(t)}{dt}$?

The function $\,\int\_0^t \ln{(s^2+t^2)} ds$ is defined to be continuous in $s^2+t^2 > 0$ and $ s^2+t^2 \ne 0$, then by the Fundamental Theorem of Integral Calculus, we have $$ f´(t) = (\int\_0^t \ln{(s^2+t^2)} ds)´= f(t) - f(0) = \ln{(s^2+t^2)} + \ln{(s^2)}$$

Is that correct? can I apply the same procedure to this function $g(t) = \int\_0^t \frac{e^{st}}{s} ds$ ? Please help. Thanks.