Linearity of indefinite integrals I am trying to make sense of 'linearity' of indefinite integrals.
Let us restrict to the 1-dimensional case. My point is that $\int 0\, dx = C \in \mathbb{R}$, so I cannot really say that $\int$ is a linear operator. Indeed, linearity of $f \colon V \rightarrow W$ ($V,W$ vector spaces) implies $f(0) = 0$. 
In order to define $\int \colon V \rightarrow W$ in a good way, I should introduce an equivalence relation $\sim$ on $W$ saying that two elements are in the same equivalence class if they coincide up to a constant. So $\int$ becomes linear as operator $\int \colon V \rightarrow W_{/\sim}$. Assume a primitive of $f$ is $F$. Then
$$\int f(x)\,dx = [F(x)] \in W_{/\sim}$$
or, as usual, $F(x)+C, C \in \mathbb{R}$. So I would say that the result of an indefinite integral is actually a coset in some quotient vector space. This even solves the problem that $\int \colon V \rightarrow W $ is not a well defined function. 
My question is: is this a good way to think, or is there a better one? I have never seen such a thing, neither in a course of Analysis, nor in the books I have read. I am wondering why. It seems a very natural thing to do when introducing indefinite integrals.
 A: Of course this is a good way to think. But we need some extra notation. Given an interval $J\subset{\mathbb R}$ call two functions $F : J\to{\mathbb R}$, $\>G:J\to{\mathbb R}\>$ equivalent if $F-G$ is constant on $J$. It is then easy to see that the equivalence classes $\langle F\rangle$ form a real vector space in the obvious way. Let $V$ be the subspace generated by the $C^1$-functions on $J$. Then
$$D:\quad V\to C^0(J), \qquad\langle F\rangle\mapsto F'$$
is a linear isomorphism with inverse the undetermined integral:
$$\int:\quad f\mapsto \int f(t)\>dt\ .$$ Thereby each "differentiation rule" generates an "integration rule" as follows:
$$F'=f\quad\Longrightarrow\quad \int f(t)\>dt=\langle F(t)\rangle\ .$$
And on and on.
A: See the section on operator notation here: https://en.wikipedia.org/wiki/Linearity_of_integration
It should answer your question. We can treat the integral as the inverse of the differential operator that takes its values in some vector space modulo the constant functions as the whole kernel of the differential operator. 
Remark: There are possibly better, or perhaps more complicated ways to think of the integral. Depending on what you need it for I guess. It sounds like you're definitely making sense of it. 
