Deceptive Integral

$\begin{aligned} I&=\int_{0}^{\tfrac{1}{2}} \dfrac{1+x^2}{(1-x^2)\sqrt{1+x^4}} dx\\ &\text{Dividing numerator and denominator by } x^2\\ I&=\int_{0}^{\tfrac{1}{2}} \dfrac{\left(1+\dfrac{1}{x^2}\right)\hspace{3mm}dx}{\left(\dfrac{1}{x}-x\right)\sqrt{\dfrac{1}{x^2}+x^2}}\\ &=- \displaystyle\int_{0}^{\tfrac{1}{2}} \dfrac{\left(1+\dfrac{1}{x^2}\right)\hspace{3mm}dx}{\left(x-\dfrac{1}{x}\right)\sqrt{\left(x-\dfrac{1}{x}\right)^2+2}}\\\\ \text{Let } u&= x-\dfrac{1}{x}\\ du&=\left(1+\dfrac{1}{x^2}\right)\hspace{3mm}dx\\ \implies I&=-\int_{-\infty}^{-\tfrac{3}{2}}\dfrac{du}{u\sqrt{u^2+2}}\\\\ \text{Let } t&= \sqrt{u^2+2}\\ dt&=\dfrac{u\hspace{3mm}du}{\sqrt{u^2+2}}\\ I&=-\int_{\infty}^{\tfrac{\sqrt{17}}{2}}\dfrac{dt}{t^2-2}\\\\ &=-\dfrac{1}{2\sqrt{2}} \ln\Biggl|\dfrac{t-\sqrt{2}}{t+\sqrt{2}}\Biggr|_{\infty}^{\tfrac{\sqrt{17}}{2}}\\ &=\dfrac{1}{2\sqrt{2}} \ln\Biggl|\dfrac{t+\sqrt{2}}{t-\sqrt{2}}\Biggr|_{\infty}^{\tfrac{\sqrt{17}}{2}}\\ &=\dfrac{1}{2\sqrt{2}}\left( \ln\Biggl|\dfrac{\tfrac{\sqrt{17}}{2}+\sqrt{2}}{\tfrac{\sqrt{17}}{2}-\sqrt{2}}\Biggr|-\lim_{t \to \infty}\ln\Biggl|\dfrac{t+\sqrt{2}}{t-\sqrt{2}}\Biggr|\right)\\ &=\dfrac{1}{2\sqrt{2}} \left( \ln\Biggl|\dfrac{\sqrt{17}+2\sqrt{2}}{\sqrt{17}-2\sqrt{2}}\Biggr|\right)\hspace{6mm}\color{#3D99F6}\small\lim_{t \to \infty}\ln\Biggl|\dfrac{t+\sqrt{2}}{t-\sqrt{2}}\Biggr|=0\\\\ &\approx\color{#EC7300}\boxed{\color{#333333}0.594217}\end{aligned}$