$\text{ The total number of distinct values of }x\in [0,1]\text{ for which }{\int }_0^x\frac{t^2}{1+t^4}dt=2x-1\text{ is }$

$Letf\left(x\right)=\underset{0}{\overset{x}{\int }}\frac{t^2}{1+t^4}dt-2x+1$

$Clearlyf\left(0\right)=1,f\left(1\right)=\underset{0}{\overset{1}{\int }}\frac{t^2}{1+t^4}dt-1$

$Now{t}^2+\frac{1}{t^2}\ge 2\Rightarrow \frac{t^2}{t^4+1}\le \frac{1}{2}$

$\therefore f\left(1\right)<0$

$Also{f}^1\left(x\right)=\frac{x^2}{1+x^4}-2$

$=\frac{1}{x^2+\frac{1}{x^2}}-2$

$Since{x}^2+\frac{1}{x^2}\ge 2$

$\Rightarrow f^1\left(x\right)<0$

$\therefore fisStrictly decrea\sin g andf\left(0\right)>0,f\left(1\right)<0$

$\therefore Exactlyonerootforf\left(x\right)in(0,1).$