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!

$$\int_0^\infty \frac{1-x^2}{(1+x^2)^2}\operatorname{sech}^2 \left(\frac{\pi x}{2}\right) dx\overset{IBP}=\pi\int_{0}^\infty \frac{x}{1+x^2}\operatorname{sech}^2 \left(\frac{\pi x}{2}\right)\tanh\left(\frac{\pi x}{2}\right)dx$$
$$=2\pi\int_{-\infty}^\infty \frac{x}{1+x^2}\color{blue}{\left(\frac{e^{\pi x/2}}{\left(e^{\pi x/2}+e^{-\pi x/2}\right)^3}-\frac{e^{-\pi x/2}}{\left(e^{\pi x/2}+e^{-\pi x/2}\right)^3}\right)}dx$$
$$\overset{x\to -x}=\color{blue}{2}\cdot 2\pi\int_{-\infty}^\infty  \frac{x}{1+x^2}\color{blue}{\frac{e^{\pi x/2}}{\left(e^{\pi x/2}+e^{-\pi x/2}\right)^3}}dx\overset{\pi x\to x}=4\pi\int_{-\infty}^\infty 
 \frac{x}{\pi^2+x^2}\frac{e^{2 x}}{(1+e^{ x})^3}dx$$
$$=4\pi\int_{-\infty}^\infty \Im\left(-\color{red}{\frac{1}{\pi+ix}}\right)\frac{e^{2 x}}{(1+e^{ x})^3}dx=-4\pi\Im\int_{-\infty}^\infty \color{red}{\int_0^\infty e^{-(\pi+ix)t}}\frac{e^{2 x}}{(1+e^{ x})^3}\color{red}{dt}dx$$
$$\small \overset{\large e^x\to x}=-4\pi\Im\int_0^\infty e^{-\pi t}\left(\lim_{a\to 1-it}\int_{0}^\infty\frac{x^a}{(1+x)^3}dx\right)dt=-4\pi\Im\int_0^\infty e^{-\pi t}\left(\lim_{a\to 1-it}\frac{\pi}{2}\frac{ a(1-a)}{\sin(\pi a)}\right)dt$$
$$=2\pi^2\int_0^\infty e^{-\pi t}\frac{t^2}{\sinh(\pi t)}dt\overset{\pi t =x}=\frac{4}{\pi} \int_0^\infty \frac{e^{-2x }x^2}{1-e^{-2x}}dx\overset{\large e^{-x}\to x}=\frac{4}{\pi}\int_0^1 \frac{x\ln^2 x}{1-x^2}dx$$
$$\overset{x^2\to x}=\frac{2}{\pi}\int_0^1\frac{\ln^2 x}{1-x}dx=\frac{2}{\pi}\sum_{n=1}^\infty \int_0^1 x^{n-1}\ln^2  x\, dx=\frac{2}{\pi}\sum_{n=1}^\infty\frac{1}{2n^3}=\frac{\zeta(3)}{\pi}$$
