From "Inverse functions and differentiation":
Integrating this relationship gives $$ f^{-1}(x)=\int\frac{1}{f'(f^{-1}(x))}\,dx + c. $$ This is only useful if the integral exists.
How can this be helpful at all to find the inverse function by integration, when the inverse itself is part of the integrand? An example would be nice...
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The formula is helpful to double check problems, where the inverse function is a mapping of the derivative of the function, like in Function whose inverse is also its derivative? or Functions whose derivative is the inverse of that function.
This seems to be extendable when the inverse function might be written in terms of the function itself or derivatives of it...
The formula will be useful if $f'(x)$ can be expressed as a function of $f(x)$. i.e there exists a function $g(y)$ such that $f'(x) = g(f(x))$. For an example, consider the case $f(x) = \tan(x)$, we know
$$\frac{d}{dx} \tan(x) = 1 + \tan(x)^2\quad\iff\quad g(y) = 1 + y^2$$ Plug this into your formula and notice $f(0) = 0$, you get
$$f^{-1}(x) = \tan^{-1}(x) = \int_0^x \frac{dt}{f'(f^{-1}(t))} =\int_0^x \frac{dt}{g(t)} = \int_0^x \frac{dt}{1+t^2} $$ For $|t| < 1$, we have $$\frac{1}{1+t^2} = \sum_{n=0}^\infty (-1)^n t^{2n} \quad\implies\quad \tan^{-1}(x) = \sum_{n=0}^\infty (-1)^n \int_0^x t^{2n} dx = \sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{2n+1} $$ The relation you have allow you to construct a power series representation of $\tan^{-1}(x)$ for $|x| < 1$.