Is this famous identity for indefinite integrals axiom or theorem? Let say that $A(x)$ describes area under $f(x)$ in the interval $[0, x]$ ($x$ varies). Then, if we sum up all infinitesimally small rectangles each of area $f(x)dx$ for all $x$-values in the domain of $f(x)$ we will get exact area under $f(x)$ from $0$ to $x$.We write this process as $ \int_{0}^{x}f(x)dx = A(x).$
But, we can also easly prove that instant velocity of $A(x)$  for input $x$ is of course $f(x)$ (that is instant change in the added area, $f(x)dx,$ over $dx$ which is indeed $f(x)$). So, $\frac{d}{dx}A(x) = f(x).$
This was for some interval, but if we want $\int_{-\infty}^{\infty}f(x)dx$ for whole number line, my intuition tells me that integral would be some function, not a number (except if x-axis is asymptote to both directions of our function, for example $e^{-x^2}$).
Now, my question.
How can we prove that $$\int_{-\infty}^{\infty}f(x)dx = A(x) + C?$$
I understand that $\frac{d}{dx} [A(x) + C] = \frac{d}{dx}A(x)$ but I can't see how can previous proven facts in calculus prove this. So, is this the definition, axiom or theorem which can be proven? Thanks

From the comments on the question, it appears you are asking for an explanation of why $\int f(x) dx= A(x)+C$. This isn't the same was what appears in the question itself, but I can't make sense of the equation $\int_{-\infty}^{\infty} dx= A(x)+C$. So my answer addresses the interpretation of your question based on the comments.
Define $A(x)=\int_0^{x} f(t) dt$. Then (part of) the Fundamental Theorem of Calculus states that $\frac{d}{dx} A(x) = f(x)$. Now, the notatation $\int f(x) dx$ denotes the family of antiderivatives of $f(x)$. We've just seen from FTC that $A(x)$ is an antiderivative of $f(x)$. If $F(x)$ is another antiderivative, then
$$
\frac{d}{dx}(A(x)-F(x))= f(x)-f(x)=0
$$
It follows from this that $A(x)-F(x)$ is a constant function (by the Mean Value Theorem for example). So $F(x)=A(x)+C$ for some constant $C$.
Conclusion: $\int f(x) dx$  is precisely $A(x)+C$ where $C$ is an arbitrary constant.
Edit: To be more precise about answering your question, I wouldn't call this statement an axiom. Rather it is a corollary of two theorems: FTC and the fact that if a function's derivative is identically 0 then the function is constant.
Edit 2: It should also be emphasized that the notation $\int_{-\infty}^{\infty} f(x) dx$ is very different from $\int f(x) dx$. While the latter represents the family of all antiderivatives of $f(x)$, the former represents the net area under the graph of $f(x)$ over $(-\infty,\infty)$ which is a real number if it exists at all. You can connect this to $A(x)$ if you want to, e.g.,
$$
\int_{-\infty}^{\infty} f(x) dx = \int_0^{\infty} f(x) dx + \int_{-\infty}^0 f(x) dx = \lim_{x\to\infty} A(x)+\lim_{x\to -\infty} A(x)
$$
But you have to be carefuly about the limits existing, and watch for $\infty-\infty$ situations, etc. But this would work for a nice function like $\frac{1}{1+x^2}$.
