$\int x^2/(x^4+1)\mathrm{dx}=\pi /(2\sqrt{2})$ with limits 0 and $\infty$

$x^2/(x^4+1)=1/2[1/(x^2+i)+1/(x^2-i)]$

The two denominators on the rhs will be zero when

$x=\pm \sqrt{-i}$; $x=\pm \sqrt{i}$ respectively, so we can expand again:

$x^2/(x^4+1)=1/4[(1/\alpha )1/(x-\alpha )-1/(x+\alpha )+(1/\beta )1/(x-\beta )-1/(x+\beta )]$

where

$\alpha =-1/\sqrt{2}+i/\sqrt{2}$

$\beta =1/\sqrt{2}+i/\sqrt{2}$

Now consider your integral but on a closed contour being a semicircle, centred on the origin, with the 'flat side' being on the real axis, and the curved side being above the real axis.

You need to convince yourself:

(i) That as the radius tends to infinity, the integral vanishes over the 'curved side'.
(ii) The integral over the 'flat side' is twice your integral as the radius -> infinity
(iii) The first and third fractions in the equation above produce simple poles enclosed within the semicircle. The residues are $1/4\alpha$, $1/4\beta$ respectively

${\int }_C{x}^2/(x^4+1)\mathrm{dx}=2{\int }_0^{\infty }{x}^2/(x^4+1)\mathrm{dx}=2\pi i\sum$

where $C$ is the semicircle contour (in the limit as its radius -> infinity) and $\sum$is the sum of the residues, i.e.

$\sum =1/4\alpha +1/4\beta =-i/2\sqrt{2}$

So

${\int }_0^{\infty }{x}^2/(x^4+1)\mathrm{dx}=\pi i.-i/2\sqrt{2}=\pi /2\sqrt{2}$

Andre