Let $g(x)=\int\limits_{0}^{x} \frac{t^{2}}{1+t^{4}} d t-(2 x)+1$
Let $I ( x )=\int\limits_{0}^{ x } \frac{ t ^{2}}{1+ t ^{4}} dt$
$=\frac{1}{2} \int\limits_{0}^{x} \frac{2 t^{2}}{1+t^{4}} d t$
$=\frac{1}{2} \int\limits_{0}^{ x }\left(\frac{ t ^{2}+1}{1+ t ^{4}}+\frac{ t ^{2}-1}{1+ t ^{4}}\right) dt$
$=\frac{1}{2} \int\limits_{0}^{x} \frac{1+\frac{1}{t^{2}}}{\left(t-\frac{1}{t}\right)^{2}+2} d t+\frac{1}{2} \int\limits_{0}^{x} \frac{\left(1-\frac{1}{t^{2}}\right) d t}{\left(t+\frac{1}{t}\right)^{2}-2}$
$I ( x )=\frac{1}{2} \frac{1}{\sqrt{2}}\left[\tan ^{-1}\left(\frac{ t -\frac{1}{ t }}{\sqrt{2}}\right)\right]_{0}^{ x }+\frac{1}{2} \times \frac{1}{2 \sqrt{2}}\left[\ell n \left|\frac{ t +\frac{1}{ t }-\sqrt{2}}{ t +\frac{1}{ t }+\sqrt{2}}\right|\right]_{0}^{ x }$
Now, $g'(x)=\frac{x^{2}}{1+x^{4}}-2$
$g'(x)=\frac{1}{\frac{1}{x^{2}}+x^{2}}-2$
since $\frac{1}{x^{2}}+x^{2} \geq 2$
$\therefore g '( x ) < 0$
So $\frac{1}{x^{2}+\frac{1}{x^{2}}} \leq \frac{1}{2}$
Now, $g (0)=1$
$\& g(1)=\int\limits_{0}^{1} \frac{t^{2}}{1+t^{4}} d t-2+1$
$=\int\limits_{0}^{1} \frac{t^{2}}{1+t^{4}} d t-1$
$=\left[0+\frac{1}{4 \sqrt{2}} \ell n \left(\frac{2-\sqrt{2}}{2+\sqrt{2}}\right)\right]-\left[\frac{1}{2 \sqrt{2}}\left(-\frac{\pi}{2}\right)+\frac{1}{4 \sqrt{2}} \ell n \left(\frac{1}{1}\right)\right]-1$
$=\frac{1}{4 \sqrt{2}} \ell n \left(\frac{2-\sqrt{2}}{2+\sqrt{2}}\right)+\frac{\pi}{4 \sqrt{2}}-1$
$g (1)=- ve$
$\therefore g (0)>0=\& g (1)<0$
$\& g '( x )<0$
$\therefore g ( x )$ has one root in $[0,1]$