Integrating $\frac{x^2}{(1+x^2)^3}$ Over the Real Line

[Fermat1]

integrate $\frac{x^2}{(1+x^2)^3}$ over the real line

 

[chisigma]

integrate $\frac{x^2}{(1+x^2)^3}$ over the real line

[sp]The function has in the upper half plane a pole of order 3 in z=i, so that is...

$\displaystyle \int_{- \infty}^{+ \infty} \frac{x^{2}}{(1 + x^{2})^{3}}\ d x = 2\ \pi i\ \lim_{z \rightarrow i} \frac{1}{2}\ \frac {d^{2}}{d z^{2}}\ \frac{z^{2}}{(z+i)^{3}} = \frac{\pi}{8}$[/sp]


Kind regards

$\chi$ $\sigma$

 

[Fermat1]

[sp]The function has in the upper half plane a pole of order 3 in z=i, so that is...

$\displaystyle \int_{- \infty}^{+ \infty} \frac{x^{2}}{(1 + x^{2})^{3}}\ d x = 2\ \pi i\ \lim_{z \rightarrow i} \frac{1}{2}\ \frac {d^{2}}{d z^{2}}\ \frac{z^{2}}{(z+i)^{3}} = \frac{\pi}{8}$[/sp]


Kind regards

$\chi$ $\sigma$

that is a way I had no really expected (although it is correct). What about a tan substitution

 

[Saitama]

integrate $\frac{x^2}{(1+x^2)^3}$ over the real line

Is this a homework problem?

Spoiler

Suspecting that this is a homework problem, I leave a hint, use $x=\tan\theta$. The resulting integral is straightforward.