Linearity of indefinite integrals

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.