Consider this integral:
$$I = \int g\left(f(x)\right)dx.$$
Assuming all regularity conditions, by inverse function theorem, $$\frac{df(x)}{dx}=\frac{1}{\left[f^{-1}\right]'\left(f(x)\right)}$$ and so the integral can be evaluated with respect to $f(x)=y$,
$$I = \int g(y)\left[f^{-1}\right]'(y)dy.$$
This says that if we know $g(y)$, the integral $I$ will depend on $f^{-1}$ acting on the function space rather than $f$ acting on the variable $x$ .
For example, we can evaluate $I$ with respect to $f^{-1}(f(x))=log(f(x))$ or $f^{-1}(f(x))=\sqrt{f(x)}$ without knowing $f$ as a function of x. All we need to know is $f^{-1}$ acting on the function space to integrate. Am I right?