Let f and g be functions. Now consider:
The differential operator is linear -- if we use the Heaviside D notation to denote this, we may extend D-1 to mean the first integral. To say that D-1 is therefore linear requires a moment to discuss the arbitrary constant of integration; D-1 would be straightforward to show linear if the arbitrary constant of integration could be set to zero.
Abstractly, we can say that D is a linear transformation from some vector space V to another one, W. We know that D(c) = 0 for any constant function c. We can by general theory (mean value theorem)identify the subspace C of V, consisting of all constant functions as the whole kernel of D. Then by linear algebra we can establish that D-1 is a well-defined linear transformation that is bijective on Im D and takes values in V/C.
That is, we treat the arbitrary constant of integration as a notation for a coset f+C; and all is well with the argument.