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.

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, $$
\boxed{\int_{0}^{\infty} \frac{d x}{\left(x^{2}+1\right)^{n}}=\frac{\pi}{2^{2 n-1}}\left(\begin{array}{c}
2 n-2 \\
n-1
\end{array}\right)}
$$
