<s>[INST] 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.
 [/INST] 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.
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