How to prove $\int_{0}^{\infty} \frac{(1-x^2) \, \text{sech}^2\left(\frac{\pi x}{2} \right)}{(1+x^2)^2}\, dx = \frac{\zeta(3)}{\pi}$? I was recently searching for interesting looking integrals. In my search, I came upon the following result:

$$ \int_{0}^{\infty} \frac{(1-x^2) \, \text{sech}^2\left(\frac{\pi x}{2} \right)}{(1+x^2)^2}\, dx = \frac{\zeta(3)}{\pi}$$

and I wanted to try and prove it.

Inspired by this answer by Jack D'Aurizio, I took the Weierstrass product for $\cosh(x)$ to obtain
$$
\cosh\left(\frac{\pi x}{2} \right) = \prod_{n \ge 1}\left(1 + \frac{x^2}{(2n-1)^2} \right)
$$
And by logarithmically differentiating twice we get
$$
\frac{\pi^2}{4}\text{sech}^2\left(\frac{\pi x}{2} \right)  = \sum_{n \ge 1} \frac{4(2n-1)^2}{\left(x^2 + (2n-1)^2\right)^2} - \frac{2}{x^2 + (2n-1)^2}
$$
Which means we get
\begin{align*}
\int_{0}^{\infty} \frac{(1-x^2) \, \text{sech}^2\left(\frac{\pi x}{2} \right)}{(1+x^2)^2}\, dx & =\frac{4}{\pi^2}\sum_{n\ge 1} \int_{0}^{\infty}  \frac{(1-x^2)}{(1+x^2)^2}\left( \frac{4(2n-1)^2}{\left(x^2 + (2n-1)^2\right)^2} - \frac{2}{x^2 + (2n-1)^2}\right)\, dx
\end{align*}
However, after this, I couldn't figure out how to evaluate the resulting integral.
Does anyone know how I could continue this method? Or alternatively, does anyone know another way in which the result can be proven? Thank you very much!!
Edit:
Per jimjim's request, I'll add that I found this integral on the Wikipedia article for $\zeta(3)$. I believe the reference is to this text where the following formula is given
$$
(s-1) \zeta(s) = 2\pi \int_{\mathbb{R}}\frac{\left(\frac{1}{2} + xi \right)^{1-s}}{\left(e^{\pi x} +e^{-\pi x} \right)^2}\, dx
$$
which for the case of $s=3$ reduces to the surprisingly concise
$$
\int_{\mathbb{R}}\frac{\text{sech}^2(\pi x)}{(1+2xi)^2} \, dx = \frac{\zeta(3)}{\pi}
$$
And I presume that one can modify the previous equation to get to the original integral from the question, but it is not apparent to me how this may be done.
Edit 2:
Random Variable has kindly posted in the comments how to go from $\int_{\mathbb{R}}\frac{\text{sech}^2(\pi x)}{(1+2xi)^2} \, dx$ to $ \int_{0}^{\infty} \frac{(1-x^2) \, \text{sech}^2\left(\frac{\pi x}{2} \right)}{(1+x^2)^2}\, dx$. Thank you very much!

After playing around a bit more with the Weierstrass product method, I realized I could proceed to do the integration with partial fractions. Using that
$$
\int_{0}^{\infty} \frac{1}{(x^2+a^2)^n} \ dx = \frac{(2n-3)!!}{(2n-2)!!}\frac{\pi}{2a^{2n-1}}, \quad a\ge0,\ n \in \mathbb{N}
$$
as shown in this answer, we get the following
$\require{cancel}$
\begin{align*}
I&  \overset{\color{blue}{n\to n-1}}{=}\frac{4}{\pi^2} \sum_{n\ge \color{blue}{0}} \int_{0}^{\infty}  \frac{(1-x^2)}{(1+x^2)^2}\left( \frac{4(2n\mathbin{\color{blue}{+}}1)^2}{\left(x^2 + (2n\mathbin{\color{blue}{+}}1)^2\right)^2} - \frac{2}{x^2 + (2n\mathbin{\color{blue}{+}}1)^2}\right)\, dx\\
& \overset{\color{purple}{a=2n+1}}{=}\frac{4}{\pi^2}\Bigg(\int_{0}^{\infty}\underbrace{\frac{8}{(x^2 +1)^4}-\frac{8}{(x^2 +1)^3} +\frac{2}{(x^2 +1)^2}}_{\color{red}{n=0}}\ dx +\sum_{n\ge \color{red}{1}} \Bigg[\frac{4(a^2 +1)}{(a^2 -1)^2}\int_{0}^{\infty}  \frac{1}{(x^2 +1)^2}\, dx \\
& \qquad- \frac{2(a^4+6a^2 +1)}{(a^2 -1)^3}\int_{0}^{\infty}  \frac{1}{x^2 +1}\, dx +\frac{2(a^4+6a^2 +1)}{(a^2 -1)^3}\int_{0}^{\infty}  \frac{1}{x^2 +a^2}\, dx\\
&\qquad+\frac{4a^2(a^2 +1)}{(a^2 -1)^2}\int_{0}^{\infty}  \frac{1}{(x^2 +a^2)^2}\, dx \Bigg]\Bigg)\\
& =\frac{4}{\pi^2} \Bigg( \frac{40\pi}{32} - \frac{24\pi}{16} +\frac{2\pi}{4}+\sum_{n\ge 1}\Bigg[\frac{4(a^2 +1)}{(a^2 -1)^2}\frac{\pi}{4} - \frac{2(a^4+6a^2 +1)}{(a^2 -1)^3}\frac{\pi}{2}+\frac{2(a^4+6a^2 +1)}{(a^2 -1)^3}\frac{\pi}{2 a }\\
&\qquad+\frac{4a^2(a^2 +1)}{(a^2 -1)^2}\frac{\pi}{4a^3}\Bigg]\Bigg)\\
& =\frac{4}{\pi} \left( \frac{1}{4}+\sum_{n\ge 1}\Bigg[\frac{a^5-a}{a(a^2 -1)^3} - \frac{a^5+6a^3 +a}{a(a^2 -1)^3}+\frac{a^4+6a^2 +1}{a(a^2 -1)^3} +\frac{a^4-1}{a(a^2 -1)^3}\Bigg]\right)\\
& =\frac{4}{\pi} \left( \frac{1}{4}+\sum_{n\ge 1}\frac{2\cancel{a}\cancel{(a-1)^3}}{\cancel{a}(a+1)^3\cancel{(a-1)^3}}\right)\\
& \overset{\color{purple}{a=2n+1}}{=}\frac{4}{\pi} \left( \frac{1}{4}+\frac{1}{4}\sum_{n\ge 1}\frac{1}{(n+1)^3}\right)\\
& =\frac{\zeta(3)}{\pi}
\end{align*}
