The total number of distinct $x \in[0,1]$ for which $\int\limits_0^x \frac{t^2}{1+t^4} d t=2 x-1$, is

1

Let $f(x)=\int\limits_0^x \frac{t^2}{1+t^4} d t-2 x+1$. Clearly, $f(x)$ is continuous and

$f'(x)=\frac{x^2}{1+x^4}-2=\frac{1}{\frac{1}{x^2}+x^2}-2=\frac{1}{\left(x-\frac{1}{x}\right)^2+2}-2<0$

So, $f(x)$ is a strictly decreasing function on $[0,1]$.

Also, $f(0)=1>0$

and, $f(1)=\int\limits_0^1 \frac{t^2}{1+t^4} d t-1<0 \quad\left[∵ \int\limits_0^1 \frac{t^2}{1+t^4} d t \leq \int\limits_0^1 \frac{1}{2} d t=\frac{1}{2}\right]$

So, $f(x)=0$ has exactly one solution in $[0,1]$.