Total number of distinct $x\in[0,1]$ for which $\int_{0}^{x}\frac{t^2}{1+t^4}dt=2x-1$ 
Find total number of distinct $x\in[0,1]$ for which $$\int_{0}^{x}\frac{t^2}{1+t^4}dt=2x-1$$

My multiple attempts are as follows:-
Attempt $1$:
$$t^2=\tan\theta$$
$$2t\dfrac{dt}{d\theta}=\sec^2\theta$$
$$dt=\dfrac{\sec^2\theta}{2\sqrt{\tan\theta}}d\theta$$
$$\int_{0}^{\tan^{-1}x^2}\dfrac{\tan\theta}{1+\tan^2\theta}\dfrac{\sec^2\theta}{2\sqrt{\tan\theta}}d\theta=2x-1$$
$$\int_{0}^{\tan^{-1}x^2}\sqrt{\tan\theta}d\theta=4x-2$$
$$\int_{0}^{\tan^{-1}x^2}\dfrac{\sqrt{\sin\theta}}{\sqrt{\cos\theta}}=4x-2$$
$$\int_{0}^{\tan^{-1}x^2}\dfrac{\sin\theta}{\sqrt{\sin\theta}\sqrt{\cos\theta}}=4x-2$$
$$\cos\theta=y$$
$$\int_{1}^{\frac{1}{\sqrt{1+x^4}}}-\dfrac{dy}{\sqrt{y}\sqrt{\sqrt{1-y^2}}}=4x-2$$
Now it is unsolvable from here.
Attempt $2$:
It seems like this integral is unsolvable, so let's apply Riemann sum
$$\lim_{h\to0}h(f(0)+f(h)+f(2h)+f(3h)\cdots\cdots f(nh))=2x-1$$
$$nh=x$$
$$\lim_{h\to0}h\sum_{r=0}^{n}f(rh)=2x-1$$
$$\lim_{h\to0}h\sum_{r=0}^{n}\dfrac{r^2h^2}{1+r^4h^4}=2x-1$$
If we put the limit $h\rightarrow 0$, then $0\cdot(0+0+0\cdots\cdots)$ where inside the bracket we have indeterminate form as $n\rightarrow \infty$
Now I was not getting how to proceed further.
 A: Let $f(x)=\int_0^{x}\frac {t^{2}} {1+t^{4}} dt-2x+1$. Let us show that $f(0) >0, f(1) <0$ and $f$ is strictly decreasing. These facts would imply that $f(x)=0$ for exactly one value of $x$.
$f(0) >0$ is obvious. $f(1)= \int_0^{1}\frac {t^{2}} {1+t^{4}}dt -1<\int_0^{1}1 dt -1=0$ (since $(t^{2}-1)^{2} \geq 0$ so $t^{4} +1\geq 2t^{2} >t^{2}$). It remains to show that $f'(x) <0$ for all $x$. Now $f'(x)=\frac {x^{2}} {1+x^{4}}-2<0$. [I will leave it to you to verify this last inequality]. 
A: Hint: Notice, that:
$$\int\frac{x^2dx}{x^4+1}=
\frac{1}{2}\int\frac{2x}{x^4+1}xdx$$
Now knowing, that $\frac{d}{dx}x=1$ and $\frac{d}{dx}\arctan{x^2}=\frac{2x}{x^4+1}$, we can solve it by parts.
A: $x \in [0,1]$;
$F(x):=\displaystyle{\int_{0}^{x}}\dfrac{t^2}{1+t^4}dt -2x+1$;
1) $F'(x)=\dfrac{x^2}{1+x^4} -2 < x^2-2<$
$1-2<0$;
Since $F$ is decreasing there is at most one zero in $[0,1]$.
2) $F(1)< \displaystyle{\int_{0}^{1}}t^2dt -1=-2/3$;
$(1/2)\displaystyle{\int_{0}^{1}}t^2dt-1=$
$(1/6)-1=-5/6<F(1)$, i.e.
$-5/6<F(1)<-2/3$.
3) $F(0)=1$.
4) Since $F$ is continuos there is one zero in $[0,1]$(IVT).
Since $F$ is strictly decreasing there is exactly one zero in $[0,1]$.
We have $F(x_0)=0$, where  $1/2 < x_0 < 1$ (Why?).
A: Actually you do not need to perform any explicit computation. $\frac{x^2}{1+x^4}=\frac{d}{dx}\int_{0}^{x}\frac{t^2}{1+t^4}\,dt$ is an increasing function over $[0,1]$ and it is bounded between $0$ and $\frac{1}{2}$, so $f(x)=\int_{0}^{x}\frac{t^2}{1+t^4}\,dt$ is a non-negative, increasing and convex function over $[0,1]$ with a derivative bounded by $\frac{1}{2}$. It follows that
$$ g(x)=\int_{0}^{x}\frac{t^2}{1+t^4}\,dt-2x+1 $$
is a decreasing and convex function over $[0,1]$. $g(0)$ and $g(1)$ have opposite signs, so $g(x)=0$ has a unique solution in $(0,1)$ (close to $0.5231289$)
