How to find the general solution to $\int f^{-1}(x){\rm d}x$ in terms of $\int f(x){\rm d}x$

I am trying to find a general proof of $\int f^{-1}(x)\,{\rm d}x$ in terms of $\int f(x)\,{\rm d}x$. The first step that I took was to piece apart what it means for a function to have an inverse. So I know the way an inverse function works, but I don’t know how it works in integration like in proving this. I am very interested in seeing the solution to this because knowing what this is would help to solve integrals where you know what the integral for the inverse function is.

Solution 1:

Note that $$\int f^{-1}(x)\,dx=\int 1\cdot f^{-1}(x)\,dx = x\cdot f^{-1}(x)-\int x\cdot (f^{-1})’(x)\,dx=x\cdot f^{-1}(x)-\int \frac{x}{f’(f^{-1}(x))}\,dx$$

Now, we calculate $\int \frac{x}{f’f^{-1}(x)}\,dx$:

Let $F=\int f(x)\,dx$, and make the substitutions of $u=f^{-1}(x)\implies x=f(u)\implies dx=f’(u)\,du$. Therefore, we have transformed our integral to:

$$\int \frac{f(u)}{f’(u)}f’(u)\,du=F(u)=F\left(f^{-1}(x)\right)$$

$$\therefore \int f^{-1}(x)\,dx = x\cdot f^{-1}(x)-F\left(f^{-1}(x)\right)$$