First, let's show

$$\int_0^{\infty}\frac{x^2}{1+x^4}dx=\int_0^{\infty}\frac{1}{1+x^4}dx$$

For this, let $t=\frac{1}{x}$. Then $dt=-\frac{1}{x^2}dx$ or $dx=-\frac{1}{t^2}dt$. It follows that

$$\int_0^{\infty}\frac{x^2}{1+x^4}dx=\int_{\infty}^0\frac{\frac{1}{t^2}}{1+\frac{1}{t^4}}\left(-\frac{1}{t^2}\right)dt=\int_0^{\infty}\frac{1}{1+t^4}dt$$

Therefore,

$$\begin{align*} \int_0^{\infty}\frac{x^2}{1+x^4}dx &= \frac{1}{2}\left( \int_0^{\infty}\frac{x^2}{1+x^4}dx +  \int_0^{\infty}\frac{1}{1+x^4}dx\right)  \\ &= \frac{1}{2} \int_0^{\infty}\frac{1+ x^2}{1+x^4}dx \\ &=\frac{1}{2}\int_0^{\infty}\frac{1+\frac{1}{x^2}}{x^2 + \frac{1}{x^2}}dx\\&=\frac{1}{2}\int_0^{\infty}\frac{1}{x^2 + \frac{1}{x^2}}d\left(x-\frac{1}{x}\right) \\&=\frac{1}{2}\int_0^{\infty}\frac{1}{\left(x-\frac{1}{x}\right)^2 + 2}d\left(x-\frac{1}{x}\right) \end{align*}$$

Let $u=x-\frac{1}{x}$, then $u\in(-\infty, \infty)$. And the above integral equals

$$\frac{1}{2}\int_{-\infty}^{\infty}\frac{1}{u^2 + 2}du =\frac{\sqrt{2}}{4} \int_{-\infty}^{\infty}\frac{1}{\left(\frac{u}{\sqrt{2}}\right)^2+1}d\left(\frac{u}{\sqrt{2}}\right)=\frac{\sqrt{2}}{4}\arctan{\frac{u}{\sqrt{2}}}|_{-\infty}^{\infty}=\boxed{\frac{\sqrt{2}\pi}{4}}$$