Suppose we find an invertible function $f(x)$ and we want to investigate properties of its inverse.
The inverse function theorem tells us, if I am interpreting correctly, $\frac{d}{dx}f^{-1}(x) = \frac{1}{\frac{d}{dx}f (x)|_{f^{-1}(x)}}$ (weird notation since stackexchange claims prime marks case double exponents).
Now, let's suppose we make a substitution, $x = f^{-1}(u),$ then, is the result
$$\frac{d}{dx}f^{-1}(x)|_{f(u)} * \frac{d}{du}f(u) = \frac{1}{\frac{d}{du}f(u)},$$
and thus I can integrate with respect to $u$ on the left and right sides to obtain an expression given an explicit $f(x)$?
My concern is the chain rule when I make the substitution and thus affecting the validity of whether or not I may integrate to obtain an explicit expression.
I think the answer is "no".
When I make the substitution, the derivative of $f^{-1}(x)$ has already been taken.
Thus, $\int \frac{d}{du}f^{-1}(u)|_{f(u)}$ is its own unique function and the integral and derivative operations do not cancel. In order for them to cancel, I would need the extra $\cdot f'(u)$ on the outside which is not there.