How to find the formula for the integral $\int_{0}^{\infty} \frac{d x}{\left(x^{2}+1\right)^{n}}$, where $n\in N$? By the generalization in my post,we are going to evaluate the integral $$\int_{0}^{\infty} \frac{d x}{\left(x^{2}+1\right)^{n}},$$
where $n\in N.$
First of all, let us define the integral $$I_n(a)=\int_{0}^{\infty} \frac{d x}{\left(x^{2}+a\right)^{n}} \textrm{ for any positive real number }a.$$
Again, we start with $$I_1(a)=\int_{0}^{\infty} \frac{d x}{x^{2}+a}= \left[\frac{1}{\sqrt{a}} \tan ^{-1}\left(\frac{x}{\sqrt{a}}\right)\right]_{0}^{\infty} = \frac{\pi}{2 }a^{-\frac{1}{2} } $$
Then differentiating $I_1(a)$ w.r.t. $a$ by $n-1$ times yields
$$
\int_{0}^{\infty} \frac{(-1)^{n-1}(n-1) !}{\left(x^{2}+a\right)^{n}} d x=\frac{\pi}{2} \left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right) \cdots\left(-\frac{2 n-3}{2}\right) a^{-\frac{2 n-1}{2}}
$$
Rearranging and simplifying gives $$
\boxed{\int_{0}^{\infty} \frac{d x}{\left(x^{2}+a\right)^{n}} =\frac{\pi a^{-\frac{2 n-1}{2}}}{2^{n}(n-1) !} \prod_{k=1}^{n-1}(2 k-1)}
$$
Putting $a=1$ gives the formula of our integral
$$
\boxed{\int_{0}^{\infty} \frac{d x}{\left(x^{2}+1\right)^{n}} =\frac{\pi}{2^{n}(n-1) !} \prod_{k=1}^{n-1}(2 k-1)= \frac{\pi}{2^{2 n-1}} \left(\begin{array}{c}
2 n-2 \\
n-1
\end{array}\right)}$$
For verification, let’s try $$
\begin{aligned}
\int_{0}^{\infty} \frac{d x}{\left(x^{2}+1\right)^{10}} &= \frac{\pi}{2^{19}}\left(\begin{array}{c}
18 \\
9
\end{array}\right) =\frac{12155 \pi}{131072} ,
\end{aligned}
$$
which is checked by Wolframalpha .
Are there any other methods to find the formula?  Alternate methods are warmly welcome.
Join me if you are interested in creating more formula for those integrals in the form $$
\int_{c}^{d} \frac{f(x)}{\left(x^{m}+1\right)^{n}} d x.
$$
where $m$ and $n$ are natural numbers.
 A: Beta Function
One approach is to use the Beta Function:
$$
\begin{align}
\int_0^\infty\frac{\mathrm{d}x}{(1+x^2)^n}
&=\frac12\int_0^\infty\frac{x^{-1/2}\,\mathrm{d}x}{(1+x)^n}\tag{1a}\\[6pt]
&=\tfrac12\operatorname{B}\left(\tfrac12,n-\tfrac12\right)\tag{1b}\\
&=\tfrac12\frac{\sqrt\pi\cdot\sqrt\pi\frac{(2n-2)!}{2^{2n-2}(n-1)!}}{(n-1)!}\tag{1c}\\[3pt]
&=\frac\pi{2^{2n-1}}\binom{2n-2}{n-1}\tag{1d}
\end{align}
$$
Explanation:
$\text{(1a)}$: substitute $x\mapsto x^{1/2}$
$\text{(1b)}$: Beta Function
$\text{(1c)}$: $\Gamma(1/2)=\sqrt\pi$, $\Gamma(n-1/2)=\sqrt\pi\frac{(2n-2)!}{2^{2n-2}(n-1)!}$, $\Gamma(n)=(n-1)!$
$\text{(1d)}$: simplify

Contour Integration
Another approach is to use contour integration. First, we compute the residue of $\frac1{\left(z^2+1\right)^n}$ at $z=i$:
$$\newcommand{\Res}{\operatorname*{Res}}
\begin{align}
\Res_{z=i}\frac1{\left(z^2+1\right)^n}
&=\Res_{z=i}\frac1{(z-i)^n}\frac1{(z+i)^n}\tag{2a}\\
&=\left[u^{-1}\right]\frac1{u^n}\frac1{(u+2i)^n}\tag{2b}\\[3pt]
&=\left[u^{-1}\right]\frac1{(2i)^n}\frac1{u^n}\frac1{\left(1+\frac{u}{2i}\right)^n}\tag{2c}\\
&=\frac1{(2i)^n}\binom{-n}{n-1}\left(\frac1{2i}\right)^{n-1}\tag{2d}\\
&=\frac{-i}{2^{2n-1}}\binom{2n-2}{n-1}\tag{2e}
\end{align}
$$
Explanation:
$\text{(2a)}$: $\frac1{z^2+1}=\frac1{z-i}\frac1{z+i}$
$\text{(2b)}$: substitute $z\mapsto u+i$; the residue at $u=0$ is the coefficient of $u^{-1}$
$\text{(2c)}$: pull out a factor of $(2i)^{-n}$ to prepare for using the Binomial Theorem
$\text{(2d)}$: $\left[u^{-1}\right]u^{-n}(1+\frac{u}{2i})^{-n}=\left[u^{n-1}\right]\left(1+\frac{u}{2i}\right)^{-n}$
$\phantom{\text{(2d):}}$ which can be computed using the Binomial Theorem
$\text{(2e)}$: simplify using negative binomial coefficients
Now it is fairly easy to compute
$$
\begin{align}
\int_0^\infty\frac{\mathrm{d}x}{(x^2+1)^n}
&=\frac12\int_\gamma\frac{\mathrm{d}z}{(z^2+1)^n}\tag{3a}\\[3pt]
&=\pi i\Res_{z=i}\frac1{\left(z^2+1\right)}\tag{3b}\\
&=\frac\pi{2^{2n-1}}\binom{2n-2}{n-1}\tag{3c}
\end{align}
$$
Explanation:
$\text{(3a)}$: use the contour $\gamma=\lim\limits_{R\to\infty}[-R,R]\cup Re^{i[0,\pi]}$
$\phantom{\text{(3a):}}$ which circles the singularity at $i$ once counter-clockwise
$\phantom{\text{(3a):}}$ the integral along the arc vanishes as $R\to\infty$
$\text{(3b)}$: Residue Theorem
$\text{(3c)}$: apply $(2)$
A: Denote $I_n = \int_0^\infty \frac1{(1+x^2)^n}dx$, and note that
$$\left( \frac x{(1+x^2)^{n-1}} \right)’
= \frac {2(n-1)}{(1+x^2)^{n}}- \frac {2n-3}{(1+x^2)^{n-1}}
$$
Integrate both sides to get the recursion $I_n= \frac{2n-3}{2(n-1)}I_{n-1} $, along with $I_1=\frac\pi2$. Thus
$$I_n = \frac{(2n-3)!(n-1)\pi }{[2^{n-1} (n-1)!]^2}$$
A: Along the line of @robjohn's solution, consider the change of variables $u=\frac{1}{1+x^2}$. Then
$$du=-2 u^2\Big(\frac{1}{u}-1\Big)^{1/2} dx=-\frac12 u^{3/2}(1-u)^{-1/2}dx$$
and so
$$\int^\infty_0\frac{dx}{(1+x^2)^n}=\frac12\int^1_0u^{n-\tfrac12-1}(1-u)^{\tfrac12-1}\,du=\frac12B\big(n-\tfrac12,\frac12\big)=\frac12\frac{\Gamma(n-\tfrac12)\Gamma(\tfrac12)}{\Gamma(n)}$$
The identity of the OP follows from properties of the Gamma function.
A: If you use complex analysis, then there is a pole at $x = i$
If we take the contour of the semi-circle in the upper half-plane, then when $n\ge 1$ the integral along the semi-circle vanishes.  That means that we can focus on the residual at $i.$
$\int_0^\infty \frac {1}{(x^2 + 1)^n} \ dx = \frac {\pi i}{(n-1)!} \frac {d^{n-1}}{dx^{n-1}} \frac {1}{(x+i)^n}$ evaluated at $i$
$\frac {\pi (2n-2)!}{((n-1)!)^2(2)^{2n-1}}$
Since it seem like you are interested:
$\int_0^\infty \frac {1}{x^n + 1}\ dx = \frac {\pi \csc \frac {\pi}{n}}{n}$
And, I am thinking that $\int_0^\infty \frac {1}{(x^m + 1)^n}\ dx$  will give something along the lines of $\frac {\pi \csc \frac {\pi}{m}}{m^n} {2n-2 \choose n-1}.$ But, I haven't worked it out.
A: Thanks to Mr Nejimban, I am now going to evaluate the integral by converting it into a Wallis integral by the substitution $x=\tan \theta$, which yields
$$I_{n}(1)= \int_{0}^{\infty} \frac{d x}{\left(x^{2}+1\right)^{n}}  =\int_{0}^{\frac{\pi}{2}} \frac{\sec ^{2} \theta d \theta}{\sec ^{2 n} \theta} =\int_{0}^{\frac{\pi}{2}} \cos ^{2 n-2} \theta d \theta
$$
By the Wallis Formula for cosine, $$
\int_{0}^{\frac{\pi}{2}} \cos ^{2 n-2} \theta d\theta =\frac{\pi}{2} \prod_{k=1}^{n-1} \frac{2 k-1}{2 k},
$$
Simplifying yields $$
\begin{aligned}
\int_{0}^{\frac{\pi}{2}} \cos ^{2 n-2} \theta &=\frac{\pi}{2} \prod_{k=1}^{n-1}\left(\frac{2 k-1}{2 k} \cdot \frac{2 k}{2 k}\right) \\
&=\frac{\pi}{2}\cdot \frac{(2 n-2) !}{2^{2 n-2}[(n-1) !]^{2}} \\
&=\frac{\pi}{2^{2 n-1}}\left(\begin{array}{c}
2 n-2 \\
n-1
\end{array}\right)
\end{aligned}
$$
$$
\int_{0}^{\infty} \frac{d x}{\left(x^{2}+a\right)^{n}} \stackrel{x \rightarrow \frac{x}{\sqrt{a}}}{=} \frac{1}{a^{n-\frac{1}{2}}} I_{n}(1)=\frac{\pi}{a^{\frac{2n-1}{2} } \cdot 2^{2 n-1}}\left(\begin{array}{c}
2 n-2 \\
n-1
\end{array}\right)
$$
In particular, $$
\boxed{\int_{0}^{\infty} \frac{d x}{\left(x^{2}+1\right)^{n}}=\frac{\pi}{2^{2 n-1}}\left(\begin{array}{c}
2 n-2 \\
n-1
\end{array}\right)}
$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
&\bbox[5px,#ffd]{\left.\int_{0}^{\infty}{\dd x \over \pars{x^{2} + 1}^{n}}
\right\vert_{\Re(n)\ >\ 1/2}}
\,\,\,\stackrel{x^{2}\ \mapsto\ x}{=}\,\,\,
{1 \over 2}\int_{0}^{\infty}{x^{\color{#f00}{1/2} - 1} \over \pars{1 + x}^{n}}\,\dd x
\end{align}

Since $\ds{\pars{1 + x}^{-n} =
\sum_{k = 0}^{\infty}{-n \choose k}x^{k} =
\sum_{k = 0}^{\infty}\bracks{{n + k - 1\choose k}\pars{-1}^{k}}x^{k} =
\sum_{k = 0}^{\infty}\color{#f00}{\Gamma\pars{n + k} \over \Gamma\pars{n}}{\pars{-x}^{k} \over k!};}$
\begin{align}
& \overbrace{\bbox[5px,#ffd]{\left.\int_{0}^{\infty}{\dd x \over \pars{x^{2} + 1}^{n}}
\right\vert_{\Re(n)\ >\ 1/2}} =
{1 \over 2}\,\Gamma\pars{1 \over 2}{\Gamma\pars{n - 1/2}\over \Gamma\pars{n}}}
^{\ds{Ramanujan's\ Master\ Theorem}}
\\[5mm] = & {\pi \over 2}{\pars{n - 3/2}! \over \pars{n - 1}!\pars{-1/2}!} =
\bbox[5px,#ffd]{{\pi \over 2}{n - 3/2 \choose n - 1}}
\end{align}
