Roots of $\int_{0}^{x} \frac{t^2}{1+t^4} dt=2x-1, x \in [0,1]$ $$\int_{0}^{x} \frac{t^2}{1+t^4} dt=2x-1, x \in [0,1]$$
This equation is known to have one real root in $[0,1]$, but when I d.w.r.t. $x$ by apploying Lebnitz rule, I get
$$\frac{x^2}{1+x^4}=2,$$
this equation has 4 non-real roots. I would like to know my mistake as I  am not getting a real root. I would line to know the correct solution of this problem.
 A: You cannot differentiate the equation simply as the equality does not hold for all $x$. Instead let $$f(x)=\int_{0}^x\frac{t^2}{1+t^4}\mathrm{d}t$$
Then,
$$f'(x)=\frac{x^2}{1+x^4}\quad, \ g(x)=2x-1 ,\quad g'(x)=2$$
consider the function $h(x)=f(x)-g(x)$ clearly  $h'(x)<0$ in $[0,1]$
observe that $h(0)=1$.  Also in $[0,1]$,  $\frac{x^2}{1+x^4}\le0.5$
or  $f(1)=\int_{0}^1\frac{t^2}{1+t^4}dt\le0.5$, thus $h(1)\le-0.5$. By mean value theorem $h(x)=0$ or $f(x)=g(x)$ for some $c$ in $[0,1]$.
Since h(x) is monotonic there will only be $1$ solution.
A: Another idea to solve it :
$x \in [0,1]\to \frac{t^2}{1+t^4}=\frac{t^2}{1-(-t^4)}=t^2(1-t^4+t^8-t^{12}+...)$ so $$\int_{0}^{x} \frac{t^2}{1+t^4} dt=2x-1\\
\int_{0}^{x} t^2(1-t^4+t^8-...) dt=2x-1\\
\frac{x^3}{3}-\frac{x^7}{7}+\frac{x^{11}}{11}-...=2x-1$$ approximate it $$\frac{x^3}{3}=2x-1 \to x=0.524..\\
\frac{x^3}{3}-\frac{x^7}{7}=2x-1 \to x=0.5231..\\
\frac{x^3}{3}-\frac{x^7}{7}+\frac{x^{11}}{11}=2x-1\to x=0.5231..$$ can you take over  ?
A: If you use the identity
$$1+t^4=\left(t^2-\sqrt{2} t+1\right) \left(t^2+\sqrt{2} t+1\right)$$ and partial fraction decomposition, you have
$$\frac {t^2}{1+t^4}=\frac{t}{2 \sqrt{2} \left(t^2-\sqrt{2} t+1\right)}-\frac{t}{2 \sqrt{2}
   \left(t^2+\sqrt{2} t+1\right)}$$ which is simple to integrate.
As a result (after recombining the logarithms),
$$f(x)=\int_{0}^{x} \frac{t^2}{1+t^4} dt=\frac1{4\sqrt 2}\left(\log \left(\frac{x^2-\sqrt{2} x+1}{x^2+\sqrt{2} x+1}\right)+2 \tan ^{-1}\left(\sqrt{2} x+1\right)-2 \tan ^{-1}\left(1-\sqrt{2} x\right)\right)$$ and now you look for the zero of function
$$g(x)=f(x)-2x+1$$ We have $$g(0)=1 \qquad \text{and} \qquad g(1)=\frac{\pi +\log \left(3-2 \sqrt{2}\right)}{4 \sqrt{2}}-1 \,\,<0$$ So, there is one root.
To get it, be lazy and use Newton method with $x_0=\frac 12$; this will give the following iterates
$$\left(
\begin{array}{cc}
n & x_n \\
 0 & 0.50000000 \\
 1 & 0.52300273 \\
 2 & 0.52312903 \\
 3 & 0.52312904
\end{array}
\right)$$
