Let be $f$ a differentiable function such that for all $x\in \Bbb R$ it is true that $$ \int_0^{f(x)} f^2(t)dt=f(f(x)).$$
Prove that exist $c\in \Bbb R$ such that $f(f(x))=(c) exp \Biggl( \int_0^{f(x)} f(t)dt))\Biggr)$
Hello! I cannot solve this problem, I tried with the derivate difinition on $g(x)=f(f(x))/ exp \Biggl( \int_0^{f(x)} f(t)dt))\Biggr)$ but I failed...
Let $F(x)=\displaystyle\int_{0}^{f(x)}f(t)dt$, then $F'(x)=f(f(x))f'(x)$. Differentiate both sides to the equation that \begin{align*} \int_{0}^{f(x)}(f(t))^{2}dt=f(f(x)), \end{align*} we get \begin{align*} -(f(f(x)))^{2}f'(x)+f'(f(x))f'(x)=0. \end{align*} Multiply $\exp(-F(x))$ to the later equation, we get \begin{align*} (\exp(-F(x))f(f(x)))'&=\exp(-F(x))(-F'(x))f(f(x))+\exp(-F(x))f'(f(x))f'(x)\\ &=\exp(-F(x))(-f(f(x))f'(x))(f(f(x)))+\exp(-F(x))f'(f(x))f'(x)\\ &=-\exp(-F(x))(f(f(x)))^{2}f'(x)+\exp(-F(x))f'(f(x))f'(x)\\ &=\exp(-F(x))(-(f(f(x)))^{2}f'(x)+f'(f(x))f'(x))\\ &=0,\\ \end{align*} so $f(f(x))=c\exp(F(x))$ for some constant $c$.