How to find $\int_0^{\infty}\frac{dx}{(1+x^2)^4}$ How would you compute for the definite integral of 
$$\int_0^{\infty}\frac{dx}{(1+x^2)^4}$$
I know that integral of $\displaystyle \frac1{(1+x^2)}$ equals $\tan^{-1}x$. I tried using integration by parts without much luck.  My teacher pointed me to special functions by which I found out about the hypergeometric distribution. Although I don't know how to apply it to this problem.
Anybody know how to use special functions or how to go about this problem?
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\begin{align}
{\cal F}\pars{\mu}&\equiv
\int_0^{\infty}{\dd x \over \mu + x^{2}}
=\mu^{-1/2}\int_0^{\infty}{\dd x \over 1 + x^{2}} = \half\,\pi\mu^{-1/2}
\end{align}
\begin{align}
{\cal F}^{'''}\pars{\mu}&\equiv
-3!\int_0^{\infty}{\dd x \over \pars{\mu + x^{2}}^{4}}
=\half\,\pi\,\totald[3]{\mu^{-1/2}}{\mu}
=\half\,\pi\,\pars{-\,\half}\pars{-\,{3 \over 2}}\pars{-\,{5 \over 2}}\mu^{-7/2}
\end{align}
Set $\mu = 1$ in both members:
$$
\color{#00f}{\large\int_0^{\infty}{\dd x \over \pars{1 + x^{2}}^{4}}}
={15\pi/16 \over 6}= \color{#00f}{\large{5 \over 32}\,\pi}
$$
A: Integration by parts does in fact help.
Let $I_n = \int (1+x^2)^{-n}dx$.
Then multiply by 1 and integrate by parts:
$$I_n = x (1+x^2)^{-n} - \int x \cdot (-n)2x (1+x^2)^{-n-1}dx = \dots = x (1+x^2)^{-n} + 2n(I_n - I_{n+1}).$$
(In the "..." step, write $x^2=(x^2+1)-1$ in the numerator and then divide.)
From this you can solve for $I_{n+1}$ in terms of $I_n$, and since you know $I_1 = \arctan x + C$, you can recursively compute $I_2$, $I_3$, $I_4$, etc.
(This is pretty much the same method that Sivaram pointed you to for finding $\int \cos^6 \theta \, d\theta$.)
A: The method given below works in general for any integral of the form $\displaystyle \int_{0}^{\infty} \frac{dx}{(1+x^2)^n}$.
Plug in $x = \tan(\theta)$. The integral becomes
$$\displaystyle \int_{0}^{\infty} \frac{dx}{(1+x^2)^4} = \int_{0}^{\pi/2} \frac{\sec^2(\theta) d \theta}{(1+\tan^2(\theta))^4} = \int_{0}^{\pi/2} \cos^{6}(\theta) d\theta = \frac{5}{6}\frac{3}{4}\frac{1}{2}\frac{\pi}{2} = \frac{5}{32}\pi$$
where the last integral has been done in a previous post here. The integration over there has been done for $\sin^n(\theta)$ but the same method works for $\cos^n(\theta)$.
If you are familiar with complex analysis, you could solve it using complex analysis by extending the function into the complex domain, choose a semicircular contour with the diameter along the real axis. Look for the poles inside the contour, (there are four poles at $z=+i$) and compute the residues. Let the radius of the semicircle tend to infinity and evaluate the integral using Cauchy residue theorem.
A: Ten years later (with Residue Calculus).
If $\gamma^R=\gamma_1^R\cup \gamma_2^R$ is the contour
$$
\gamma_R^1(t)=t, \,\,\,t\in [-R,R], \quad
\gamma_R^2(t)=R\mathrm{e}^{it}, \,\,\,t\in [0,\pi],
$$
then
$$
\int_{\gamma_R}\frac{dz}{(1+z^2)^n}=2\pi i\,
\mathrm{Res}\bigg(\frac{1}{(1+z^2)^n},z=i\bigg).
$$
But
$$
\bigg|\int_{\gamma_R^2}\frac{dz}{(1+z^2)^n}\,\bigg|=
\bigg|\int_0^\pi\frac{iR\mathrm{e}^{it}dt}{(1+\big(R\mathrm{e}^{it})\big)^n}
\bigg|\le \int_0^\pi \frac{Rdt}{(R^2-1)^n}=\frac{\pi R}{R^2-1)^n}\to 0,
$$
as $R\to\infty$, while
$$
\lim_{R\to\infty}\int_{\gamma_R^1}\frac{dz}{(1+z^2)^n}=\int_{-\infty}^\infty\frac{dx}{(1+x^2)^n}.
$$
Hence $\int_{-\infty}^\infty\frac{dx}{(1+x^2)^n}=2\pi i\,
\mathrm{Res}\bigg(\frac{1}{(1+z^2)^n},z=i\bigg).$
We use the following fact.
Fact. If $f$ has a pole of order $m$ at $z=a$, then
$$
\mathrm{Res}(f,a)=\frac{1}{(m-1)!} \left((z-a)^mf(z)\right)^{(m-1)}_{z=a}
$$
This is straightforward since
$$
f(z)=a_{-m}(z-a)^{-m}+\cdots+a_{-1}(z-a)^{-1}+a_0+a_{1}(z-a)^{1}+\cdots
$$
and hence
$$
(z-a)^mf(z)=a_m+a_{-m+1}(z-a)^{1}\cdots+a_{-1}(z-a)^{m-1}+a_0(z-a)^{m}+\cdots
$$
and thus
$$
\big((z-a)^mf(z)\big)^{(m-1)}=(m-1)!a_{-1}+m!a_0(z-a)^1+\cdots
$$
and finally $\,\,\frac{1}{(m-1)!} \big((z-a)^mf(z)\big)^{(m-1)}_{z=a}=a_{-1}$.
In our case $f(z)=\frac{1}{(1+z^2)^n}$ has a pole of order $n$, and hence
$$
\mathrm{Res}\bigg(\frac{1}{(1+z^2)^n},z=i\bigg)=
\frac{1}{(n-1)!}
\left(
(z-i)^n \frac{1}{(1+z^2)^n}
\right)^{(n-1)}_{z=i}=\frac{1}{(n-1)!}
\left(\frac{1}{(z+i)^n}
\right)^{(n-1)}_{z=i} \\
=\frac{1}{(n-1)!}\cdot (-1)^{n-1}\cdot n\cdot(n+1)\cdots(2n-2)\cdot
\left(\frac{1}{(z+i)^{2n-1}}\right)_{z=i}\\
=(-1)^{n-1}\binom{2n-2}{n-1}\cdot\frac{1}{(2i)^{2n-1}}=\frac{-i}{2^{2n-1}}\binom{2n-2}{n-1}
$$
Finally
$$
\int_{-\infty}^\infty\frac{dx}{(1+x^2)^n}=2\pi i\,
\mathrm{Res}\bigg(\frac{1}{(1+z^2)^n},z=i\bigg)=2\pi i\,\frac{-i}{2^{2n-1}}\binom{2n-2}{n-1}=\frac{\pi}{2^{2n-2}}\binom{2n-2}{n-1}.
$$
