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.

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
$$I_n = \frac{(2n-3)!(n-1)\pi }{[2^{n-1} (n-1)!]^2}$$
