Say that $u$ is a function of both $x$ and $t$; $u=u(x,t)$
Is $\displaystyle \int \frac{\partial u}{\partial t} dt = \int \frac{ du}{d t} dt$ ?
Obviously $\displaystyle \int \frac{ du}{d t} dt= u$ ($+$ constant)
$\displaystyle \int \frac{\partial u}{\partial t} dt= \frac{d}{dt} \displaystyle \int u$ $ dt$ $=u +$(constant) by the fundamental theorem of calculus?
I'm going to have to disagree there: $$ \frac{du}{dt} = \frac{\partial u}{\partial t} + \frac{\partial u}{\partial x} \frac{dx}{dt}. $$
This notation is really unhelpful, because you're using the same integration dummy variable in the derivative. I suppose what you mean is $$ \frac{d}{dt} \int_a^t u(x,s) \, ds = u(x,t), $$ by the FToC.
On the other hand, the other derivative is $$ \frac{d}{dx} \int_a^b u(x,s) \, ds = \int_a^b \frac{\partial u}{\partial x}(x,s) \, ds ,$$ by "differentiation under the integral sign", provided $a$ and $b$ are constant.
(The moral is that thinking about antiderivatives of functions of more than one variable is a bad idea: you should always have limits on your integrals, and you need to know the relationship between the variables. This becomes particularly important when you learn about the method of characteristics for PDEs, for example.)