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.

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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}
