Evaluate $\int_{-\infty}^\infty \frac{1}{(x^2+1)^3} dx$ Evaluate $\int_{-\infty}^\infty \frac{1}{(x^2+1)^3} dx$ I wasnt exactly sure how to approach this. I saw some similar examples that used Cauchy's theorem.
 A: $$\int_{-\infty}^{+\infty}\frac{dx}{(x^2+1)^3} = 2\pi i\cdot\operatorname{Res}\left(\frac{1}{(z^2+1)^3},z=i\right)=\pi i\left.\frac{d^2}{dz^2}\frac{(z-i)^3}{(z^2+1)^3}\right|_{z=i}=\color{red}{\frac{3\pi}{8}}. $$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\begin{align}&\color{#66f}{\large%
\int_{-\infty}^{\infty}{\dd x \over \pars{x^{2} + 1}^{3}}}
=\left.\totald[2]{}{\mu}\int_{0}^{\infty}{\dd x \over x^{2} + \mu}
\right\vert_{\, \mu\ =\ 1}
=\left.\totald[2]{}{\mu}\pars{\mu^{-1/2}\int_{0}^{\infty}{\dd x \over x^{2} + 1}}
\right\vert_{\, \mu\ =\ 1}
\\[5mm]&={\pi \over 2}\,\left.\totald[2]{\mu^{-1/2}}{\mu}\right\vert_{\, \mu\ =\ 1}
={\pi \over 2}\,\pars{{3 \over 4}\,\mu^{-5/2}}_{\, \mu\ =\ 1}
=\color{#66f}{\large{3\pi \over 8}}
\end{align}
A: Another way is a trig substitution $x = \tan u$ which reduces the integrand to
$$
\int \frac{dx}{(x^2+1)^3}
 = \int \frac{\sec u \tan u}{\sec^6 u} du
 = \int \sin u \cos^4 u du
 = \frac{\cos^5 u}{5}
$$
and the rest is arithmetic
A: $x=\tan{u}\rightarrow \frac{dx}{1+x^2}=du$
So the problem reduces to
$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos^4{u}\space du$
