Find all continuous function $f(x)$ defined real fro all real $x$ such that:
$$[f(x)]^2 = \int_{0}^{x} \frac{t f(t)}{1+t^2}dt$$
I am really stuck. Don't we have to differentiate both sides with respect to $t$?
$$2[f(x) f'(x)]\frac{dx}{dt} = \frac{xf(x)}{1+x^2}$$
Not sure how to proceed ...
Assume that $f$ is differentiable. Then instead of differentiating with respect to $t$, we can differentiate with respect to $x$. We get $$2f(x)f'(x)=\frac{xf(x)}{1+x^2},$$ which implies either $f(x)=0$ or $f'(x)=\dfrac{x}{2(x^2+1)}$. The latter case of course implies $f(x)=\frac14\log(x^2+1)$. These are the only two solutions.
Note that it's not possible for $f(x)$ to be $\frac 14\log(x^2+1)$ for some $x$ and $0$ elsewhere (except for $x=0$), because if $f(a)=0$ for $a> 0$, then $\int_0^a\frac{tf(t)}{1+t^2}dt>0$ but $f(a)^2=0$, a contradiction (a similar argument applies for $a<0$).
