number of distinct solution of Integral equation

$$ x\mapsto \frac{x^2}{1+x^4} $$ is increasing on $[0,1]$, hence $f(x)=\int_{0}^{x}\frac{t^2}{1+t^4}\,dt $ is non-negative and convex on $[0,1]$ and $f'(x)\leq f'(1)=\frac{1}{2}$ holds. It follows that $g(x)=f(x)-2x+1$ is decreasing on $[0,1]$. Since $g(0)=1>0$ and $$ g(1) = \frac{\pi}{4\sqrt{2}}-\frac{\log(1+\sqrt{2})}{2\sqrt{2}}-1 <0$$ it follows that $g(x)=0$ has only one solution in $[0,1]$.

Just to find out the only solution whose existence has been given by Jack.

Integrating $$\displaystyle \int_{0}^{x}\frac{t^2}{1+t^4}dt=\frac{1}{4\sqrt 2}\left(\log \left(\frac{x^2-\sqrt 2 x+1}{x^2+\sqrt 2 x+1}\right)+2\arctan(1+\sqrt2)-2\arctan(1-\sqrt2)\right) $$ Now graphing $f(x)=\displaystyle \int_{0}^{x}\frac{t^2}{1+t^4}dt$ without the denominator $4\sqrt2$ and $g(x)=4\sqrt2(2x-1)$ we get

In which the only point of intersection is approximately $(x,y)=(0.523,0.262)$ hence $$\color{red}{x\approx 0.523}$$