Integrating the derivative of a multivariate function If $u=u(x,t)$ how do I compute $\displaystyle \int \frac{du}{dt}dt$ ? 

Would I be correct in saying it is not simply a case of cancelling the $dt$'s and getting an answer of $u$+ constant, seeing as $u$ is a multivariate function?
What does the integration mean?

Integration with one variable $t$ is an operation acts on a function with the variable $t$

So $\int \frac{d}{dt}u(x,t) dt$ means nothing more than we integrate the function $\frac{d}{dt}u(x,t)$ with respect to $t$ with $x$ fixed. If $x$ was a function  of a variable $t$, then it must have been noted before one uses the integrations. However, it's not the case.

Hence, it is just $F(x)=\int \frac{d}{dt}u(x,t) dt$. And to emphasize that $x$ is fixed under the integration, set $u_x(t):=u(x,t)$

Then, $F(x)=\int \frac{d}{dt}u_x(t) dt = u_x(t) + C = u(x,t)+ C$.

This is just the fundamental theorem of calculus.

I don't have time to write more so I hope you understand it now.
\begin{equation}
\frac{du}{dt}dt=du
\end{equation}
So
\begin{equation}
\int \frac{du}{dt}dt= \int du
\end{equation}
Where
\begin{equation}
du=u_{x}dx+u_{t}dt
\end{equation}
Where
$$u_{x_{k}}=\frac{\partial u}{\partial x_{k}}$$