The total number of distinct  $x\in [0,1]$  for which $\underset{0}{\overset{x}{\int }}\frac{t^2}{1+t^4}dt=2x-1$  is

 $\underset{0}{\overset{x}{\int }}\frac{t^2}{1+t^4}dt=2x-1$
Let   $f\left(x\right)=\underset{0}{\overset{x}{\int }}\frac{t^2}{1+t^4}dt-2x+1$
$\Rightarrow f\text{'}\left(x\right)=\frac{x^2}{1+x^4}-2<0\forall x\in [0,1]$
Now, $f\left(0\right)=1$  and  $f\left(1\right)=\underset{0}{\overset{1}{\int }}\frac{t^2}{1+t^4}dt-1$
As   $0\le \frac{t^2}{1+t^4}<\frac{1}{2}\forall t\in [0,1]$
$\Rightarrow \underset{0}{\overset{1}{\int }}\frac{t^2}{1+t^4}dt<\frac{1}{2}\Rightarrow f\left(1\right)<0$
$\Rightarrow f\left(x\right)=0$ has exactly one root in [0, 1].