<s>[INST] How many methods are there to evaluate $\int_0^{\infty} \frac{1}{\left(x+\frac{1}{x}\right)^{2n}}$? Background
As I had found the integral
$$I=\int_0^{\infty} \frac{1}{\left(x+\frac{1}{x}\right)^2} d x =\frac{\pi}{4}, $$
by using $x\mapsto \frac{1}{x}$ yields
$\displaystyle I=\int_0^{\infty} \frac{\frac{1}{x^2}}{\left(\frac{1}{x}+x\right)^2} d x\tag*{} $
Averaging them gives the exact value of the integral
$\displaystyle \begin{aligned}I & =\frac{1}{2} \int_0^{\infty} \frac{1+\frac{1}{x^2}}{\left(x+\frac{1}{x}\right)^2} d x \\& =\frac{1}{2} \int_0^{\infty} \frac{d\left(x-\frac{1}{x}\right)}{\left(x-\frac{1}{x}\right)^2+4} d x \\& =\frac{1}{4}\left[\tan ^{-1}\left(\frac{x-\frac{1}{x}}{2}\right)\right]_0^{\infty} \\& =\frac{1}{4}\left[\frac{\pi}{2}-\left(-\frac{\pi}{2}\right)\right] \\& =\frac{\pi}{4}\end{aligned}\tag*{} $

I guess that we can similarly evaluate the general integral
$$
I_n=\int_0^{\infty} \frac{d x}{\left(x+\frac{1}{x}\right)^{2 n}}
$$
by mapping $x\mapsto \frac{1}{x}$ and then averaging.
$$
I_n=\frac{1}{2} \int_0^{\infty} \frac{1+\frac{1}{x^2}}{\left(x+\frac{1}{x}\right)^{2 n}} d x=\frac{1}{2} \int_0^{\infty} \frac{d\left(x-\frac{1}{x}\right)}{\left[\left(x-\frac{1}{x}\right)^2+4\right]^n}
$$
Letting $x-\frac{1}{x}=\tan \theta$ yields
$$
\begin{aligned}
I_n & =\frac{1}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{2 \sec ^2 \theta d \theta}{4^n \sec ^{2 n} \theta} =\frac{1}{4^n} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos ^{2 n-2} \theta d \theta= \boxed{\frac{\pi(2 n-3) ! !}{4^n(2 n-2) ! !}}
\end{aligned}
$$
where the last answer comes from the Wallis cosine formula.
My question: How many methods are there to evaluate $\int_0^{\infty} \frac{1}{\left(x+\frac{1}{x}\right)^{2n}}$?
 [/INST] Writing the integral on the form
$$
I_n = \int_0^{\infty} \frac{x^{2n}}{(x^2+1)^{2n}}dx
$$
can be seen as a Mellin transform of $f(x)=(1+x)^{-\rho}$. The Mellin transform is given by
$$
\mathcal{M}[f(x);s]=\frac{1}{\Gamma(\rho)}\Gamma[s,\rho-s]$$
Which then gives that
$$
 \mathcal{M}[f(x^2);s]=\frac{1}{2\Gamma(\rho)}\Gamma[s/2,\rho-s/2]
$$
With $\rho=2n$, $s=2n+1$, this gives
$$
I_n = \frac{1}{2\Gamma(2n)}\Gamma[n+1/2,n-1/2] = \frac{1}{2}B(n+1/2,n-1/2)
$$
</s>