How to show $\int\frac{d}{dx}(a^u)dx=a^u+C$ more rigorously? We all know that $$\int\frac{d}{dx}(a^u)dx=a^u+C$$ where I am differentiating with respect to $x$. But how can I write it in a more rigorous way like for example using the Fundamental Theorem of Calculus or any other method?
How can I deduce that $$\int d(a^u)=a^u+C$$
The notation looks a bit odd for me because normally we view integral as the antiderivative or the limit of Riemann sum, but this time what does $\int d(a^u)$ actually mean?
Also, when are we allowed to swap the order of differentiation and integration?
Sorry, maybe I need to clarify that $a$ is a constant and $u$ is a function of $x$.
Just for reference, my knowledge about differentials is that $dy=f'(x)dx$.
Thanks for the explanation.

In calculus (handling of derivatives and primitives, etc.) it is accepted to present functions $f:\ x\mapsto f(x)$  as function terms $f(x)$. This is in contrast to "abstract analysis" where $f(x)$ just denotes the value of the function $f$ at the point $x\in{\rm dom}(f)$, and is a number.
Without saying it you are actually considering a function  $x\mapsto u(x)$ and then  the function $f(x):=a^{u(x)}$. Therefore the term
$${d\over dx}a^u$$
appearing in your question means the function $x\mapsto f'(x)$ obtained by using the chain rule on $x\mapsto f(x)=a^{u(x)}$. 
The expression
$$\int{d\over dx}a^{u(x)}\>dx=\int f'(x)\>dx$$
by definition denotes the set of all primitives of $f'(x)$, and this is the set of all functions $x\mapsto f(x)+C$, $\>C\in{\mathbb R}$. As $f(x)=a^{u(x)}$  we therefore have
$$\int{d\over dx}a^{u(x)}\>dx=a^{u(x)}+C\ ,$$
as stated in your question. Apart from the "dangling" $C$ appearing in formulas of this sort there is no "more rigorous way" of describing the situation at hand. In particular the FTC has nothing to do with this. The only "integral theorem" we have tacitly used is the following: A function whose derivative is $\equiv0$ on some interval $I\subset{\mathbb R}$ is constant on $I$.
