Find $f$ such that $\int_{x}^{f^{-1}(x)}f(t)dt=x^2$ for all $x\in \mathbb{R}.$ Here $f$ is a bijection of $\mathbb{R}$ in $\mathbb{R}$ and is differentiable with derivative not equal to $0.$
I tried to apply the fundamental theorem of calculus but I get $$x(f^{-1}(x))'-f(x)=2x.$$ But I am not being able to get rid of $f^{-1}(x)$ in order to get a differential equation in terms of $f(x).$ I know that $(f^{-1}(x))'=\frac{1}{f'(f^{-1}(x))}$ But this does not help either.
Try this function: $f(x)=(-1+\sqrt{2})x$. I was looking for a solution in the form of $f(x)=cx$, then $f^{-1}(x)=\frac{x}{c}$ and $\int_{x}^{f^{-1}(x)}f(t)dt=\int_{x}^{x/c}ct \:dt= \frac{c}{2}(\frac{x^2}{c^2}-x^2)=\frac{cx^2}{2}(\frac{1}{c^2}-1)=x^2$. Thus, we have an equation for $c$: $\frac{c(1-c^2)}{c^2}=2$, $1-c^2=2c$.