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!

1)Integral evaluation applying Differentiation Under the Integral Sign:
First, let's consider the following function:
$$f( \alpha ) =\int _{0}^{\infty }\frac{1-x^{2}}{\left( 1+x^{2}\right)^{2}}\frac{\tanh( \alpha x)}{x} dx$$
Then similarly to OP's series expansion for $\operatorname{sech}^2\left(\frac{\pi x}{2}\right)$, let's employ the series expansion of $\tanh(\alpha x)$ presented in Section 1.421  of Gradshteyn and Ryzhik
$$f( \alpha )=\frac{2}{a}\sum _{n\geqslant 1}\int _{0}^{\infty }\frac{1-x^{2}}{\left[ x^{2} +\left(\frac{\pi ( 2n-1)}{2a}\right)^{2}\right]\left( 1+x^{2}\right)^{2}} dx$$
This integral is pretty straight foward and just requires partial fractions:
$$\begin{align} &\int _{0}^{\infty }\frac{1-x^{2}}{\left( x^{2} +\beta ^{2}\right)\left( 1+x^{2}\right)^{2}} dx \\\ &= \int _{0}^{\infty }\left[\frac{\beta ^{2} +1}{\left( \beta ^{2} -1\right)^{2}\left( x^{2} +\beta ^{2}\right)} -\frac{\beta ^{2} +1}{\left( \beta ^{2} -1\right)^{2}\left( 1+x^{2}\right)} +\frac{2}{\left( \beta ^{2} -1\right)\left( 1+x^{2}\right)^{2}}\right] dx\\\ &=\left[\frac{\left( \beta ^{2} +1\right)\arctan\left(\frac{x}{\beta }\right)}{\beta \left( \beta ^{2} -1\right)^{2}} -\frac{\left( \beta ^{2} +1\right)\arctan( x)}{\left( \beta ^{2} -1\right)^{2}} +\frac{\frac{x}{1+x^{2}} +\arctan( x)}{\left( \beta ^{2} -1\right)}\right]_{0}^{\infty }\\\ &=\frac{\pi }{2\beta ( \beta +1)^{2}}\end{align}$$
Then
$$\begin{align}f(\alpha)=\frac{\pi }{a}\sum _{n\geqslant 1}\frac{1}{\frac{\pi ( 2n-1)}{2a}\left( 1+\frac{\pi ( 2n-1)}{2a}\right)^{2}} &=\sum _{n\geqslant 1}\left[\frac{1}{n-\frac{1}{2}} -\frac{1}{n-\frac{1}{2} +\frac{a}{\pi }} -\frac{a/\pi }{\left( n-\frac{1}{2} +\frac{a}{\pi }\right)^{2}}\right]\\&=\psi ^{( 0)}\left(\frac{a}{\pi } +\frac{1}{2}\right) -\psi ^{( 0)}\left(\frac{1}{2}\right) -\frac{a}{\pi } \psi ^{( 1)}\left(\frac{a}{\pi } +\frac{1}{2}\right)\end{align}$$
$\require{cancel}$
Differentiating and setting $\alpha=\frac{\pi}{2}$
$$\begin{align}\frac{d}{d\alpha}f\left(\frac{\pi}{2}\right) &=\int _{0}^{\infty }\frac{1-x^{2}}{\left( 1+x^{2}\right)^{2}}\operatorname{sech}^2\left(\frac{\pi x}{2}\right) dx\\&=\cancel{{\frac{1}{\pi } \psi ^{( 1)}( 1)} }-\cancel{\frac{1}{\pi } \psi ^{( 1)}( 1)} -\frac{1}{2\pi } \psi ^{( 2)}( 1)\\&=\frac{\zeta(3)}{\pi}\end{align}$$
2) Completing OP's solution:
To prove that the arc contribution vanishes as the raidius of the contour approaches infinity, let's apply the Estimation Lemma.
$$\begin{align}\left|\int _{\Gamma } f( z) dz\right| &\leqslant \int _{0}^{\pi }\left|\frac{1-\left( Re^{it}\right)^{2}}{\left( 1+\left( Re^{it}\right)^{2}\right)^{2}}\operatorname{sech}^{2}\left(\frac{\pi Re^{it}}{2}\right)\right|\left| Rie^{it} dt\right|\\&\leqslant \frac{R^3}{R^4}\int _{0}^{\pi }\frac{dt}{\left|\operatorname{sech}^{2}\left(\frac{\pi Re^{it}}{2}\right)\right|}\leqslant \frac{R^3}{R^4}\int _{0}^{\pi }dt\leqslant \frac{\pi}{R}\end{align}$$
Which means that:
$$\lim_{R\rightarrow \infty}\left|\int _{\Gamma } f( z) dz\right|\leqslant \lim_{R\rightarrow \infty}\frac{\pi}{R}\leqslant 0$$
