Proof that every function of a symbolic form $A$ has an antiderivative with a unique symbolic form $A'$ I'm just learning about antiderivatives. I'm curious as to how to show that for any function having a particular symbolic structure there exist only one antiderivative corresponding to that symbolic structure. 
For example: to show that there is a unique relationship between all functions of the form $f'(x)=x^{n}$ and their antiderivatives $f(x)=\frac{x^{n+1}}{n+1}+C$. That is to say: if you could invent some sort of higher level form of function which takes a function as an input and gives the derivative of the function as an output, the function should be one to one for each symbolic class. 
Another example: with $\ln x$ as the antiderivative of $\frac{1}{x}$, how can we show $\ln x$ is the only function satisfying that relationship. 
Answering this question is beyond my current mathematical abilities. I can show that a certain antiderivative belongs to a function by differentiating the antiderivative and equating it to the function. However to show that this operation is one to one over symbolic classes I have no idea where to begin. 
 A: It depends what you mean by symbolic classes.  There can be many different symbolic forms for the same function; to pick a trivial example, the antiderivative of $x^n$ can be written as $\frac{x^{n+1}}{n+1}+C,$ $\frac{2x^{n+1}}{2n+2}+C,$ $\frac{x^{n+1}}{n+\sin^2 x+\cos^2 x}+C,$ and infinitely many other ways.
In fact, every so often, somebody posts an antiderivative that they've found that's apparently quite different from what Wolfram Alpha gives as the answer, but in fact the two are equivalent.
If your question is in what sense is the antiderivative unique up to an additive constant, here's how you can look at that: If $f$ and $g$ are continuous on the real line (or on an open interval) and if $f'=g',$ then $(f-g)'=0,$ so $f-g$ is a constant function.
So a continuous antiderivative is a unique function up to an additive constant, but there's not a unique symbolic expression for it.

By the way, if the domain of the function is not a connected set (for example, if the domain is the union of two disjoint open intervals), then things don't work out as neatly.  This is essentially because you can have different additive constants on the different connected parts (even if the original function can be expressed by the same formula over the entire domain).
A: Fundamental theorem of calculus:
$$f(x) = \int_a^x f' + C.$$
