Even though integrating numerically gives $0$, we use the Dirac delta $\delta(x)$sifting property to potentially invert functions:
$$\int_{x-c}^{x+c}f(t)\delta(t-x)dt=\int_{-c}^cf(t+x)\delta(t)dt=f(x)\mathop\implies^{t\to f(t)} f^{-1}(x)= \int_{f^{-1}(x-c)}^{f^{-1}(x+c)}t\delta(f(t)-x)df(t),f^{-1}(x-c)<f^{-1}(x)<f^{-1}(x+c),c>0$$
Then rename the integral bounds:
$$f^{-1}(x)=\int_{a}^bt\delta(f(t)-x)df(t),a<f^{-1}(x)<b\tag 1$$
Confirmed here with the inverse error function. However, this integral reminds one of a floor function integral:
$$xe^x=1\implies x=\frac34-\int_0^\frac34\lfloor te^t \rfloor dt=\frac 34-\int_x^\frac34 dt$$
$\lfloor te^t \rfloor:$
which is trivial. A possible transformation is using a limit expansion of $\delta(x)$ which numerically works:
$$f^{-1}(x)=\lim_{c\to0}\frac c\pi\int_a^b \frac{t\,d f(t)}{|f(t)-x+ic|^2}$$
Does $(1)$ provide a trivial expression of the inverse function or how applicable is it to solving equations?