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How many solutions exist for following equation $$\int_0 ^x \frac {t^2}{1+t^4}dt =2x-1$$ in $[0,1]$?

I took everything on one side and thus assumed it as $f (x)=...$ and applied Newton-Leibniz Theorem to get $f'(x)=\frac {-2x^4+x^2-2}{1+x^4} $ which is negative in $[0,1]$. Now $f (0)>0$ and $f(1)<0$ so there are odd solutions to this equation. But my book gives answer as $1$. How to prove that it has exactly 1 solution? Thanks!

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Hint. Is the function $$f(x):=\int_0^x\frac {t^2}{1+t^4}dt -2x+1$$ strictly monotone in $[0,1]$? Now, as you noted, for $x\in[0,1]$ $$f'(x)=\frac {x^2}{1+x^4}-2\leq x^2-2\leq 1-2=-1<0.$$

P.S. Note that $f(0)=1$ and $$f(1)=\int_0^1\frac{t^2}{1+t^4}dt -2+1<\int_0^1 t^2dt-1=-\frac{2}{3}<0$$ so by continuity there is at least a solution in $(0,1)$.

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