Consider a differentiable function $f$ that is strictly increasing and positive. Numerical examples seem to show that:
\begin{equation} \int_{x_1}^{x_2}\frac{1}{f^{-1}(x)}dx=\int_{f^{-1}(x_1)}^{f^{-1}(x_2)}\frac{f^{\prime}(y)}{y}dy, \end{equation} where $f^{-1}$ denotes the inverse function of $f$ and $f^{\prime}(x)=\frac{\partial}{\partial x}f(x)$.
I was unable to show this and could not find any reference, so that by now I wonder whether this equality holds at all or if the examples I checked were just lucky coincidences.
Any hints would be appreciated. Thank you very much in advance.
Yes, the given formula holds. As regards your last comment, the answer is no, making the substitution $y=f^{-1}(x)$, we don't find $$\int_{x_1}^{x_2}\frac{1}{f^{-1}(x)}dx=\int_{f^{-1}(x_1)}^{f^{-1}(x_2)}\frac{y}{f^{\prime}(y)}dy.$$ By letting $y=f^{-1}(x)$ then $f(y)=x$ and $f'(y)\,dy=dx$. Moreover, since $f$ is increasing, the same holds for $f^{-1}$ and $[x_1,x_2]=[f^{-1}(x_1),f^{-1}(x_2)]$. Hence $$\int_{x_1}^{x_2}\frac{1}{f^{-1}(x)}dx=\int_{f^{-1}(x_1)}^{f^{-1}(x_2)}\frac{1}{y}(f'(y)\,dy).$$ You may also obtain the same result by using the substitution rule $$\int_{x_1}^{x_2}g(\varphi(x)) \varphi'(x)\,dx=\int_{\varphi(x_1)}^{\varphi(x_1)}g(y)\,dy$$ where $$g(x)=\frac{f'(x)}{x}=\frac{1}{x\cdot (f^{-1})'(f(x))}\quad\text{and}\quad\varphi(x)=f^{-1}(x).$$