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

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SUMMARY

The integral of the function $\frac{x^2}{(1+x^2)^3}$ over the real line is evaluated using complex analysis techniques. Specifically, the integral is computed as $\int_{-\infty}^{+\infty} \frac{x^2}{(1+x^2)^3}\, dx = \frac{\pi}{8}$. This result is derived by identifying a pole of order 3 at $z=i$ in the upper half-plane and applying the residue theorem. The discussion also briefly mentions the possibility of using a tangent substitution for integration.

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

$\chi$ $\sigma$
 
chisigma said:
[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
 
Fermat said:
integrate $\frac{x^2}{(1+x^2)^3}$ over the real line

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

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