1. Let $F(x)=\int_0^x \frac{t^2}{1+t^4}\,dt.$ We need the number of distinct $x\in[0,1]$ satisfying $F(x)=2x-1.$

2. Define $G(x)=F(x)-(2x-1)=\int_0^x \frac{t^2}{1+t^4}\,dt-2x+1.$ We must count the number of zeros of $G(x)$ in $[0,1]$.

3. First check the endpoint values.

At $x=0$: $G(0)=0-0+1=1>0.$

At $x=1$: $G(1)=\int_0^1 \frac{t^2}{1+t^4}\,dt-1.$ Now for $t\in[0,1]$, $0\le \frac{t^2}{1+t^4}\le t^2,$ so $\int_0^1 \frac{t^2}{1+t^4}\,dt \le \int_0^1 t^2\,dt=\frac13.$ Hence $G(1)\le \frac13-1=-\frac23<0.$ So $G(0)>0$ and $G(1)<0$, which guarantees at least one root in $(0,1)$ by the Intermediate Value Theorem.

4. Now prove uniqueness. Differentiate $G(x)$: $G'(x)=\frac{x^2}{1+x^4}-2.$ For $x\in[0,1]$, $\frac{x^2}{1+x^4}\le x^2\le 1,$ therefore $G'(x)\le 1-2=-1<0.$ So $G(x)$ is strictly decreasing on $[0,1]$.

A strictly decreasing continuous function can cross zero at most once. Since we already showed it crosses zero at least once, it crosses exactly once.

5. Therefore, the total number of distinct solutions is $\boxed{1}.$