Q:

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

