How do I solve $\int_0^\infty \frac{x^2}{x^4+1} \, dx$? 
$$\int_0^\infty \frac{x^2}{x^4+1} \; dx $$

All I know this integral must be solved with beta function, but how do I come to the form $$\beta (x,y)=\int_0^1 t^{x-1}(1-t)^{y-1}\;dt \text{ ?}$$
 A: By definition $$B(p, q) = \int_0^1x^{p-1}(1-x)^{q-1}\,dx.$$ By using shift $x = \frac{t}{1 + t}$ previous integral becomes $$B(p,q) = \int_0^{\infty}\frac{t^{p-1}}{(1 + t)^{p + q}}\,dt.$$
Thus we get that
\begin{align}
I & = \int_0^\infty \frac{x^2}{1 + x^4} \, dx = \{x^4 = t, dx = t^{\frac14 - 1} \, dt, x^2 = t^{\frac12}\} = \frac14\int_0^\infty \frac{t^{\frac12}t^{\frac14 - 1}}{1+t} \, dt \\[10pt]
&  = \frac14\int_0^\infty \frac{t^{\frac34 - 1}}{(1 + t)^{\frac34 + \frac14}} \, dt = \frac14 B\left(\frac14, 1 - \frac14\right) = \frac14 \frac{\Gamma(\frac14)\Gamma(\frac14 - 1)}{\Gamma(\frac14 + 1 - \frac14)} = \frac14 \frac{\Gamma(\frac14)\Gamma(\frac14 - 1)}1 \\[10pt]
& = \frac14\frac{\pi}{\sin(\frac14\pi)} = \frac{\pi}{2\sqrt2}.
\end{align}
I hope I have not made any mistakes
A: Hint: If you have to use Beta function, let $x^2 = \tan \theta$ and use the definition of Beta function in terms of sines and cosines.
A: note that $$x^4+1=x^4+2x^2+1-2x^2=(x^2+1)^2-2x^2=(x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1)$$
A: $$\begin{eqnarray*}I=\int_0^{+\infty}\frac{x^2\,dx}{x^4+1}\stackrel{\text{parity}}{=} \frac{1}{2} \int_{-\infty}^{+\infty}\frac{dx}{x^2+\frac{1}{x^2}}&=&\frac{1}{2} \int_{-\infty}^{+\infty}\frac{dx}{\left(x-\frac{1}{x}\right)^2+2} \\[8pt]
(\text{by Glasser's Master Theorem})&=&\int_0^{+\infty}\frac{dx}{x^2+2}=\color{red}{\frac{\pi}{2\sqrt{2}}}.\end{eqnarray*} $$
As a reference, http://mathworld.wolfram.com/GlassersMasterTheorem.html.
Through the Beta function, by setting $\frac{1}{x^4+1}=u$,
$$ I=\frac{1}{4}\int_{0}^{1}u^{-3/4}(1-u)^{-1/4}\,du=\frac{\Gamma\left(\frac{1}{4}\right)\,\Gamma\left(\frac{3}{4}\right)}{4}=\frac{\pi}{4\sin\frac{\pi}{4}} =\color{red}{\frac{\pi}{2\sqrt{2}}}$$
by the reflection formula for the $\Gamma$ function.
