The derivative operator $\frac{d}{dx}(f(x))$ is a map from function space to function space. So is the definite integral $\int_0^xf(x)dx$, which maps $f(x)$ to the function that gives the signed area between $f(x)$ and the $x$ axis, up to the value of $x$ that you put in.
However, $\int f(x)dx$ does not map to a specific function. An infinite amount of functions satisfy it, namely $C+\int_0^xf(x)dx$ for all $C\in \mathbb R$.
So what kind of operator is $\int dx$ ? Is it even an operator? What kind of object is it?
Edit: Because I am not entirely sure what kind of object the indefinite integral is, I'm also somewhat unsure about the exact meaning of indefinite integral equations.
For example, based on the chain rule, we can derive: $$\frac{\delta f(g(x))}{\delta x}= g'(x)f'(g(x)) \implies \int\frac{\delta f}{\delta g(x)}(g(x))dg=\int\frac{\delta g}{\delta x}(x)\frac{\delta f}{\delta g}(g(x))dx$$
Can be considered an operator with range in a quotient space.
If $C+\int_0^xf\in F$, with $F$ some function space that contains the constant functions, the real numbers can be considered a subspace of $F$ and $\int f(x)\,dx$ is a class of equivalence in the space $F/\Bbb R$.