Integral $\int_0^\infty\frac{\tanh^2(x)}{x^2}dx$ It appears that
$$\int_0^\infty\frac{\tanh^2(x)}{x^2}dx\stackrel{\color{gray}?}=\frac{14\,\zeta(3)}{\pi^2}.\tag1$$
(so far I have about $1000$ decimal digits to confirm that).
After changing variable $x=-\tfrac12\ln z$, it takes an equivalent form
$$\int_0^1\frac{(1-z)^2}{z\,(1+z)^2 \ln^2z}dz\stackrel{\color{gray}?}=\frac{7\,\zeta(3)}{\pi^2}.\tag2$$
Quick lookup in Gradshteyn—Ryzhik and Prudnikov et al. did not find this integral, and it also is returned unevaluated by Mathematica and Maple. How can we prove this result? Am I overlooking anything trivial?
Further questions: Is it possible to generalize it and find a closed form of 
$$\mathcal A(a)=\int_0^\infty\frac{\tanh(x)\tanh(ax)}{x^2}dx,\tag3$$
or at least of a particular case with $a=2$?
Can we generalize it to higher powers
$$\mathcal B(n)=\int_0^\infty\left(\frac{\tanh(x)}x\right)^ndx?\tag4$$

Thanks to nospoon's comment below, we know that 
$$\mathcal B(3)=\frac{186\,\zeta(5)}{\pi^4}-\frac{7\,\zeta(3)}{\pi^2}\tag5$$
I checked higher powers for this pattern, and, indeed, it appears that
$$\begin{align}&\mathcal B(4)\stackrel{\color{gray}?}=-\frac{496\,\zeta(5)}{3\,\pi^4}+\frac{2540\,\zeta(7)}{\pi^6}\\
&\mathcal B(5)\stackrel{\color{gray}?}=\frac{31\,\zeta(5)}{\pi^4}-\frac{3175\,\zeta(7)}{\pi^6}+\frac{35770\,\zeta(9)}{\pi^8}\\
&\mathcal B(6)\stackrel{\color{gray}?}=\frac{5842\,\zeta(7)}{5\,\pi^6}-\frac{57232\,\zeta(9)}{\pi^8}+\frac{515844\,\zeta(11)}{\pi^{10}}\end{align}\tag6$$
 A: Note
$$I=\int_{-\infty}^\infty\frac{\tanh^2 x}{2x^2}dx
\overset{t=1/e^{2x}}=\int_0^\infty \left(\frac{1-t}{1+t}\right)^2\frac{dt}{t\ln^2t}
 \overset{ibp}=4\int_0^\infty \frac{t-1}{(1+t)^3}\frac{dt}{\ln t}
$$
Let $J(a) =4 \int_0^\infty \frac{t^a-1}{(1+t)^3}\frac{dt}{\ln t}$. Then,
$J’(a)= 4\int_0^\infty \frac{t^a}{(1+t)^3}dt 
=\frac{2\pi a(1-a) }{\sin\pi a}
$ and
$$I=\int_0^1J’(a)da
\overset{ibp}=\int_0^1(4a-2)\ln \tan \frac{\pi a}2 da
\overset{t=\tan\frac{\pi a}2}=
\frac4{\pi^2}\int_0^\infty \frac{\ln t\tan^{-1}\sqrt t}{\sqrt t(1+t)}dt
$$
Let $K(a)=\int_0^\infty \frac{\ln t\tan^{-1}a\sqrt t}{\sqrt t(1+t)}dt$. Then,
$K’(a)= \int_0^\infty \frac{\ln t}{(1+t)(1+a^2t)}dt
= \frac{2\ln^2a}{1-a^2}
$ and
$$I = \frac4{\pi^2}\int_0^1 K’(a)da
=\frac8{\pi^2}\int_0^1 \frac{\ln^2a}{1-a^2} da
= \frac8{\pi^2}\cdot \frac74\zeta(3)
=\frac{14 \zeta(3)}{\pi^2}
$$
A: $$\mathcal{A}(m,n)=\int_{0}^{\infty}\frac{\tanh(z)^m}{z^n}\text{d}z\qquad m+1>n>1$$
We have
$$\begin{aligned}
&\mathcal{A}\left(1,p+1\right)
=\frac{(2^{p+1}-1)}{\pi^p\cos\frac{p\pi}{2} } \zeta(p+1)\\
&\mathcal{A}\left(2,p+1\right)=
\frac{(p+1)(2^{p+2}-1)}{\pi^{p+1}\sin\frac{p\pi}{2} } \zeta(p+2)\\
&\mathcal{A}\left(3,p+1\right)=
\frac{(2^{p+1}-1)}{\pi^p\cos\frac{p\pi}{2} } \zeta(p+1)
-\frac{(p+1)(p+2)(2^{p+3}-1)}{2\pi^{p+2}\cos\frac{p\pi}{2} }\zeta(p+3)\\
&\mathcal{A}\left(4,p+1\right)=
\frac{4(p+1)(2^{p+2}-1)}{3\cdot\pi^{p+1}\sin\frac{p\pi}{2} } \zeta(p+2)
-\frac{(p+1)(p+2)(p+3)(2^{p+4}-1)}{6\cdot\pi^{p+3}\sin\frac{p\pi}{2} }\zeta(p+4)\\
&\mathcal{A}\left(5,p+1\right)=
\frac{(2^{p+1}-1)}{\pi^p\cos\frac{p\pi}{2} } \zeta(p+1)
-\frac{5(p+1)(p+2)(2^{p+3}-1)}{6\cdot\pi^{p+2}\cos\frac{p\pi}{2} } \zeta(p+3)
+\frac{(p+1)(p+2)(p+3)(p+4)(2^{p+5}-1)}{24\cdot\pi^{p+4}\cos\frac{p\pi}{2} }\zeta(p+5)\\
&\mathcal{A}\left(6,p+1\right)=
\frac{23(p+1)(2^{p+2}-1)}{15\cdot\pi^{p+1}\sin\frac{p\pi}{2} } \zeta(p+2)
-\frac{(p+1)(p+2)(p+3)(2^{p+4}-1)}{3\cdot\pi^{p+3}\sin\frac{p\pi}{2} } \zeta(p+4)
+\frac{(p+1)(p+2)(p+3)(p+4)(p+5)(2^{p+6}-1)}{120\cdot\pi^{p+5}\sin\frac{p\pi}{2} }\zeta(p+6)
\end{aligned}$$
And substitute different p(can be a non-integer),your cases will done.

Proof:
For $n+m=2k$,consider the function
$$
f(z)=\frac{\tanh(z)^m}{z^n}
$$
And I get
$$
\int_{0}^{\infty}\frac{\tanh(z)^m}{z^n} \text{d}z
=-\pi i\sum_{k=-\infty}^{\infty} \text{Res}(f,\frac{\pi}{2}(2k+1)i )
$$
For $n+m=2k+1$,consider
$$
f(z)=\frac{\tanh(z)^m}{z^n}\ln z
$$
And I get
$$
\int_{0}^{\infty}\frac{\tanh(z)^m}{z^n} \text{d}z
=-\sum_{k=-\infty}^{\infty} \text{Res}(f,\frac{\pi}{2}(2k+1)i )
$$
I dont have time to write a full answer.Thanks.
(I will have exams for three days tomorrow.)

If we consider the function
$$
f(z)=\frac{\tanh z}{z^{1+p}}
$$
And use the keyhole contour,we can obtain
$$
(1-e^{-2p\pi i})\int_{0}^{\infty} \frac{\tanh z}{z^{1+p}}\text{d}z
= 2\pi i\sum_{k=-\infty}^{\infty}\text{Res}\left (\frac{\tanh z}{z^{1+p}}, \frac{\pi}{2}(2k+1)i  \right )
$$

A few examples
$$
\begin{aligned}
&\int_{0}^{\infty} \frac{\tanh z}{z^{3/2}} \text{d}z
=\frac{4-\sqrt{2} }{\sqrt{\pi} } \zeta\left ( \frac{3}{2}  \right )\\
&\int_{0}^{\infty} \frac{\tanh z}{z^{5/3}} \text{d}z
=\frac{2(2^{5/3}-1)}{\pi^{2/3}}\zeta\left ( \frac{5}{3}  \right )\\
&\int_{0}^{\infty} \frac{\tanh^2 z}{z^{2}} \text{d}z
= \frac{14}{\pi^2}\zeta(3)\\
&\int_{0}^{\infty} \frac{\tanh^2 z}{z^{3/2}} \text{d}z
= \frac{24-3\sqrt{2} }{2\pi^{3/2}} \zeta\left ( \frac{5}{2}  \right )\\
&\int_{0}^{\infty} \frac{\tanh^2 z}{z^{5/2}} \text{d}z
= \frac{80-5\sqrt{2} }{2\pi^{5/2}} \zeta\left ( \frac{7}{2}  \right )\\
&
\end{aligned}
$$
A: Using the Taylor series for $\tan\left(z+\frac{(2k+1)\pi}2\right)=-\frac1z+\frac z3+O\!\left(z^3\right)$ and $i\tanh(z)=\tan(iz)$, we get
$$\newcommand{\Res}{\operatorname*{Res}}
\tanh^2\left(z+i\frac{(2k+1)\pi}2\right)=\frac1{z^2}+\frac23+O\!\left(z^2\right)\tag1
$$
Using the Taylor series for $\frac1{1+z}=1-z+O\!\left(z^2\right)$, we get
$$
\frac1{\left(z+i\frac{(2k+1)\pi}2\right)^2}=-\frac4{(2k+1)^2\pi^2}-\frac{16iz}{(2k+1)^3\pi^3}+O\!\left(z^2\right)\tag2
$$
Therefore, with $z_k=i\frac{(2k+1)\pi}2$ we get that
$$
\Res\limits_{z=z_k}\left(\frac{\tanh^2(z)}{z^2}\right)=-\frac{16i}{(2k+1)^3\pi^3}\tag3
$$
We can use the contour of integration

$$
\begin{align}
\int_0^\infty\frac{\tanh^2(x)}{x^2}\,\mathrm{d}x
&=\frac12\int_{-\infty}^\infty\frac{\tanh^2(x)}{x^2}\,\mathrm{d}x\tag{4a}\\
&=\pi i\sum_{k=0}^\infty\frac{-16i}{(2k+1)^3\pi^3}\tag{4b}\\
&=\frac{14}{\pi^2}\zeta(3)\tag{4c}
\end{align}
$$
Explanation:
$\text{(4a)}$: use symmetry
$\text{(4b)}$: the integral along the contour is $2\pi i$ times the sum of the residues inside
$\text{(4c)}$: $\sum\limits_{k=0}^\infty\frac1{(2k+1)^3}=\frac78\zeta(3)$

The answer for $(3)$ from the question, with $a=2$, is given in this answer.
A: Let $\psi_{1}(z)$ be the trigamma function.
Similar to the answer here, we can integrate the function $$g(z) = \psi_{1}\left(\frac{z}{\pi i } \right) \tanh^{2}(z) $$ around an rectangular contour with vertices at $\pm R, \pm R + \pi i $.
The function $\psi_{1}\left(\frac{z}{\pi i} \right) $ has a double pole at the origin, but it's canceled by the double zero of $\tanh^{2}(z)$.
Also, as $\Re(z) \to \pm \infty$, the magnitude of $\tanh^{2}(z)$ tends to $1$, while the magnitude of $\psi_{1}\left(\frac{z}{\pi i} \right) $ decays like $\frac{\pi}{z}$.   This, coupled with the fact the height of the contour is fixed, causes the integral to vanish on the vertical sides of the contour as $R \to \infty$.
We therefore get
$$\begin{align}\lim_{R \to \infty}\oint g(z) \, \mathrm dz &=\int_{-\infty}^{\infty} \psi_{1}\left(\frac{x}{\pi i } \right) \tanh^{2}(x) \, \mathrm dx - \int_{-\infty}^{\infty}\psi_{1}\left(\frac{x}{\pi i} + 1 \right) \tanh^{2}(x) \, \mathrm dx \\ &= - \pi^{2} \int_{-\infty}^{\infty} \frac{\tanh^{2}(x)}{x^{2}} \, \mathrm dx \tag{1} \\ &= 2 \pi i \operatorname{Res}\left[g(z), \frac{\pi i}{2} \right] \\ &= 2 \pi i \left(\frac{\psi_{2} \left(\frac{1}{2} \right)}{\pi i}\right)\\& = -28 \zeta(3) \tag{2}.\end{align}$$
The result then follows.

The fastest way to determine the residue at $z = \frac{\pi i}{2}$ is to multiply the Laurent series expansion $$\tanh^{2}(z) = \frac{1}{\left(z-\frac{\pi  i}{2}\right)^{2}} + \mathcal{O}(1)$$ with the Taylor expansion of $\psi_{1} \left(\frac{z}{\pi i} \right)$ at $z= \frac{\pi i}{2}$.

$(1)$ https://en.wikipedia.org/wiki/Polygamma_function#Recurrence_relation
$(2)$ https://mathworld.wolfram.com/PolygammaFunction.html (16)

To evaluate the integral $$\mathcal A(3)= \int_{0}^{\infty} \frac{\tanh (x) \tanh(3x)}{x^{2}} \, \mathrm dx,$$ we can integrate the function $$g(z) = \psi_{1} \left(\frac{z}{\pi i} \right) \tanh (z) \tanh(3z) $$ around the same contour.
We get $$\begin{align} \mathcal{A}(3) &= \frac{1}{\pi i } \left(\operatorname{Res}\left[g(z), \frac{\pi i }{6}\right] + \operatorname{Res}\left[g(z), \frac{5 \pi i}{6}\right] + \operatorname{Res}\left[g(z), \frac{\pi i}{2} \right]\right)  \\ &= \frac{1}{\pi i} \left(\frac{i}{3} \,  \psi_{1} \left(\frac{1}{6} \right) \tan \left(\frac{\pi}{6} \right)  + \frac{i}{3} \, \psi_{1} \left(\frac{5 }{6} \right) \tan \left(\frac{5 \pi }{6} \right)- \frac{i}{3 \pi} \, \psi_{2} \left(\frac{1}{2} \right)\right) \\ &= \frac{1}{3 \sqrt{3} \, \pi} \left(\psi_{1} \left(\frac{1}{6} \right) - \psi_{1} \left(\frac{5}{6} \right) + \frac{14\sqrt{3}}{\pi} \,  \zeta(3) \right). \end{align} $$

To show that $$\operatorname{Res} \left[g(z), \frac{\pi i}{2} \right] = \frac{1}{3 \pi i} \, \psi_{2} \left(\frac{1}{2} \right),$$ we can use the formula for the residue of a double pole derived here.
$$\begin{align} \operatorname{Res} \left[g(z), \frac{\pi i}{2} \right] &= \operatorname{Res} \left[\frac{\psi_{1} \left(\frac{z}{\pi i }\right) \sinh(z) \sinh(3z)}{\cosh(z) \cosh(3z)}, \frac{\pi i}{2}  \right] \\ &=  \operatorname{Res} \left[\frac{\psi_{1} \left(\frac{z}{\pi i }\right) \left(\cosh(4z) - \cosh(2z) \right)}{\cosh(2z) +\cosh(4z)}, \frac{\pi i}{2}  \right]  \\ &= \frac{6 \left( \frac{2}{\pi i} \, \psi_{2} \left(\frac{1}{2} \right)(12) \right) - 2 \left( 2 \,  \psi_{1} \left(\frac{1}{2} \right)(0) \right)}{3(12)^{2}} \\ &= \frac{1}{3 \pi i} \, \psi_{2} \left(\frac{1}{2} \right) \end{align}$$
A: The solution for $n=3$ can be easily generalized to any $n\ge 2$: it suffices to use parity to extend the integration to $\mathbb R$ and then compute integral by residues by moving the contour to $i\infty$. The residues come from the poles of $\tanh^n z$ given by $z_k=i\pi\left(k+\frac12\right)$, $k\in \mathbb Z$.
For example, as $z\to z_k$, we have
$$\frac{\tanh^2 z}{z^2}=\frac{1}{z_k^2}\frac{1}{(z-z_k)^2}-\frac{16i}{\pi^3(2k+1)^3}\frac{1}{z-z_k}+\mathrm{reg.},$$
and therefore the integral is given by
$$\int_0^{\infty}\frac{\tanh^2 z}{z^2} dz=\frac12\cdot 2\pi i\sum_{k=0}^{\infty}\left(-\frac{16i}{\pi^3(2k+1)^3}\right)=\frac{14\zeta(3)}{\pi^2}.$$
For general $n$, we will obviously have a finite sum of zeta values.

The case of $$\mathcal I:=\int_0^{\infty}\frac{\tanh z\tanh 2 z}{z^2} dz$$ can be treated analogously. Potential poles of the integrand are given by $z^{I}_k=\frac{i\pi}{2}\left(k+\frac12\right)$ and $z^{II}_k=i\pi\left(k+\frac12\right)$, and we have
$$
\frac{\tanh z\tanh 2z}{z^2}=
\begin{cases}\frac{\tanh z_k^I}{2\left(z_k^{I}\right)^2}\frac{1}{z-z_k^I}+\mathrm{reg.}
& \text{as } z\to z_k^I,\\
\mathrm{reg.}
& \text{as } z\to z_k^{II},
\end{cases}
$$
so that the actual poles are only given by $z_k^I$.
Therefore
$$\mathcal I=\pi i\sum_{k=0}^{\infty}\frac{\tanh z_k^I}{2\left(z_k^{I}\right)^2}
=\pi i\sum_{k=0}^{\infty}\frac{i(-1)^k}{2\left(\frac{i\pi}{2}\left(k+\frac12\right)\right)^2}=\frac{8K}{\pi},$$
where $K$ denotes Catalan's constant.
A: I found a way to evaluate this integral without complex analysis, although I think that it is not rigorous yet, because I do not know how to justify the swapping of the integrals and the swapping of the integral and the infinite sum.
Anyway, start  with the identity $\displaystyle \,\,\int_0^{\infty} \frac{\sin(z x)}{\sinh(\frac{\pi}{2}x)}dx=\tanh z\,\,\,\,\,\,$ (see here for a proof, for example.)
So $$I=\int_0^{\infty} \frac{\tanh^2 z}{z^2}dz=\int_0^{\infty}\int_0^{\infty}\int_0^{\infty}\frac{\sin(zx)\sin(zt)}{z^2\sinh(\frac{\pi}{2}x)\sinh(\frac{\pi}{2}t)}dxdtdz
\\\\=\int_0^{\infty}\int_0^{\infty}\frac{f(x,t)}{\sinh(\frac{\pi}{2}x)\sinh(\frac{\pi}{2}t)}dxdt$$
Where $$f(x,t)=\int_0^{\infty}\frac{\sin(z x)\sin(x t)}{z^2}dz=\large\begin{cases} \frac{\pi}{2}x &:&0<x\le t \\\\\frac{\pi}{2}t&:&0<t\le x\end{cases}$$
which I saw on this Wikipedia list. I could not evaluate this myself, nor find a reference anywhere. (Edit: found a reference.)
Anyway, due to the symmetry, 
$$
I=\int_0^{\infty}\int_0^{\infty}\frac{f(x,t)}{\sinh(\frac{\pi}{2}x)\sinh(\frac{\pi}{2}t)}dxdt
\\=\int_0^{\infty}\int_0^{t}\frac{f(x,t)}{\sinh(\frac{\pi}{2}x)\sinh(\frac{\pi}{2}t)}dxdt+\int_0^{\infty}\int_t^{\infty}\frac{f(x,t)}{\sinh(\frac{\pi}{2}x)\sinh(\frac{\pi}{2}t)}dxdt
\\\\=2\int_0^{\infty}\int_t^{\infty}\frac{f(x,t)}{\sinh(\frac{\pi}{2}x)\sinh(\frac{\pi}{2}t)}dxdt
\\\\=2\int_0^{\infty}\int_t^{\infty}\frac{\frac{\pi}{2} t}{\sinh(\frac{\pi}{2}x)\sinh(\frac{\pi}{2}t)}dxdt
\\\\=-2\int_0^{\infty} \frac{t}{\sinh(\frac{\pi}{2} t)}\ln\tanh(\frac{\pi}{4} t)\,dt.
$$
Since $\displaystyle \,\,\,\int \frac1{\sinh x}dx=\ln\tanh(\frac{x}{2})+C.$
Now substitute $x=\tanh(\frac{\pi}{4}t)$ to get
$$I=-2\int_0^{\infty} \frac{t}{\sinh(\frac{\pi}{2} t)}\ln\tanh(\frac{\pi}{4} t)\,dt
\\=-\frac{16}{\pi^2}\int_0^1\frac{\ln x}{x}\,\operatorname{arctanh} x\, dx
\\=-\frac{16}{\pi^2}\int_0^1 \frac{\ln x}{x} \sum_{n=0}^{\infty} \frac{x^{2n+1}}{2n+1}\,dx
\\=\frac{16}{\pi^2}\sum_{n=0}^{\infty} \frac1{(2n+1)^3}
\\=\frac{14\zeta(3)}{\pi^2}.
$$
I guess similar reasoning could be used to calculate $\mathcal B(n)$ for bigger natural $n$'s.
Also, here is a reference for other evaluations of $\mathcal A(2)$.
Edit
Yet another solution.
Again, start with the identity $\displaystyle \,\,\int_0^{\infty} \frac{\sin(z x)}{\sinh(\frac{\pi}{2}x)}dx=\tanh z\,.$
Differentiate to obtain
$\displaystyle \,\int_0^{\infty} \frac{x \cos(z x)}{\sinh(\frac{\pi}{2}x)}dx=\operatorname{sech}^2 z.$
It follows that $\displaystyle \,\,\,\,\int_0^{\infty} \frac{x(1- \cos(z x))}{\sinh(\frac{\pi}{2}x)}dx=1-\operatorname{sech}^2 z=\tanh^2 z.$
Recalling that $\int_0^{\infty} \frac{\sin^2 z}{z^2}dz=\frac{\pi}{2}$, we have 
$\displaystyle \int_0^{\infty} \frac{1-\cos(z x)}{z^2}dz=\int_0^{\infty} \frac{2\sin^2(z x /2)}{z^2}dz=\frac{\pi x}{2}.$
Finally, 
$$\int_0^{\infty} \frac{\tanh^2 z}{z^2}dz=\int_0^{\infty} \int_0^{\infty}  \frac{x(1- \cos(z x))}{z^2 \sinh(\frac{\pi}{2}x)}dx dz\\
=\frac{\pi}{2}\int_0^{\infty} \frac{x^2}{\sinh(\frac{\pi}{2}x)}dx\\
=\frac{8}{\pi^2} \int_0^{\infty} x^2 \sum_{n=0}^{\infty} e^{-x(2n+1)}\,dx\\
=\frac{14\zeta(3)}{\pi^2}.$$
A: We can use an integral representation of the Dirichlet eta function to show that $$\int_{0}^{\infty} \frac{\tanh^{2}(x)}{x^{2}} \, dx = \int_{0}^{\infty} \left(1-\frac{1}{\cosh^{2}(x)} \right) \frac{dx}{x^{2}} = -56 \, \zeta'(-2) = \frac{14 \, \zeta(3)}{\pi^{2}}. $$

An integral representation of the Dirichlet eta function is  $$\eta(s) = \frac{1}{\Gamma(s)}\int_{0}^{\infty} \frac{x^{s-1}}{e^{x}+1} \, dx \, , \quad \text{Re}(s) >0. $$
Integrating by parts, we get $$ \begin{align} \eta(s) &= \frac{1}{s \, \Gamma(s)} \int_{0}^{\infty} \frac{x^{s} e^{x}}{(e^{x}+1)^{2}} \, dx= \frac{1}{4 \, \Gamma(s+1)}\int_{0}^{\infty} \frac{x^{s}}{\cosh^{2} \left(\frac{x}{2} \right)} \, dx \\ &=\frac{2^{s-1}}{\,\Gamma(s+1)} \int_{0}^{\infty} \frac{u^{s}}{\cosh^{2}(u)} \, dx \, , \quad \text{Re}(s) >-1. \tag{1} \end{align} $$
Combining $(1)$ with the Laplace transform of $x^{s}$, we get
$$\int_{0}^{\infty} \left(e^{-ax} - \frac{1}{\cosh^{2}(x)}\right)x^{s} = \Gamma(s+1) \left(\frac{1}{a^{s+1}} - \frac{\eta(s)}{2^{s-1}} \right) \, ,  \tag{2} $$ which holds for $ \text{Re}(a) >0$  and  $\text{Re}(s) >-2$.
If we restrict $s$ to the strip $-2 < \text{Re}(s) <-1$, then $(2)$ also holds for $a=0$.
So letting $s$ tend to $-2$ (and using the fact that $\eta(s)$ has a zero at $s=-2$),  we get $$ \begin{align} \int_{0}^{\infty} \left(1-\frac{1}{\cosh^{2}(x)}\right)\frac{dx}{x^{2}} &= -\lim_{s \downarrow -2} \frac{\Gamma(s+1) \eta(s)}{2^{s-1}}  \\ &= -\lim_{s \downarrow -2} \left(- \frac{1}{s+2} + \mathcal{O}(1) \right) \eta(s)  \left(8 + \mathcal{O}(s+2) \right) \\ &= 8 \lim_{s \downarrow -2} \frac{\eta(s) }{s+2} \\ &= 8 \,  \eta'(-2) \\ &= -8 \left(1-2^{3} \right) \zeta'(-2) \tag{3} \\ &=-56 \, \zeta'(-2). \end{align}$$
But by differentiating both sides of the functional equation for the Riemann zeta function, we see that $$ \zeta'(-2) = 2^{s-1} \pi^{s} \cos \left(\frac{\pi s}{2} \right) \Gamma(1-s) \zeta(1-s) \Big|_{s=-2} = -\frac{ \zeta(3)}{4\pi^{2}}.$$
Therefore, $$\int_{0}^{\infty} \left(1-\frac{1}{\cosh^{2}(x)} \right) \frac{dx}{x^{2}} = \frac{14 \zeta(3)}{\pi^{2}}. $$

$(3)$ http://mathworld.wolfram.com/DirichletEtaFunction.html (11)
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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$\ds{\large\mbox{Identity}\ \pars{1}\ \mbox{in the OP post:}}$
Integrating twice by parts:
\begin{align}
\int_{0}^{\infty}{\tanh^{2}\pars{x}  \over x^{2}}\,\dd x & =
2\int_{0}^{\infty}{\tanh\pars{x}\,\mrm{sech}^{2}\pars{x} \over x}\,\dd x
\\[5mm] & =
-6\
\underbrace{\int_{0}^{\infty}\ln\pars{x}\,\mrm{sech}^{4}\pars{x}\,\dd x}
_{\ds{\mc{J}_{4}'\pars{0}}}\ +\
4\
\underbrace{\int_{0}^{\infty}\ln\pars{x}\,\mrm{sech}^{2}\pars{x}\,\dd x}
_{\ds{\mc{J}_{2}'\pars{0}}}\label{1}\tag{1}
\end{align}

\begin{align}
\mc{J}_{\nu}\pars{\mu} & \equiv
\int_{0}^{\infty}x^{\mu}\,\mrm{sech}^{\nu}\pars{x}\,\dd x =
2^{\nu}\int_{0}^{\infty}x^{\mu}\expo{-\nu x}\,\pars{1 +\expo{-2x}}^{-\nu}\,\dd x \\[5mm] = &\
2^{\nu}\sum_{n = 0}^{\infty}{-\nu \choose n}\int_{0}^{\infty}x^{\mu}
\expo{-\pars{2n + \nu}x}\,\dd x =
2^{\nu}\,\Gamma\pars{\mu + 1}\sum_{n = 0}^{\infty}
{\nu + n - 1 \choose \nu - 1}{\pars{-1}^{n} \over \pars{2n + \nu}^{\,\mu + 1}}
\\[5mm] & =
2^{\nu - \mu - 1}\,\Gamma\pars{\mu + 1}\sum_{n = 0}^{\infty}
{\nu + n - 1 \choose \nu - 1}{\pars{-1}^{n} \over \pars{n + \nu/2}^{\,\mu + 1}}
\end{align}

\begin{align}
\mc{J}_{2}\pars{\mu} & =
2^{1 - \mu}\,\Gamma\pars{\mu + 1}\sum_{n = 0}^{\infty}
{\pars{-1}^{n} \over \pars{n + 1}^{\,\mu}} =
2^{2 - 2\mu}\pars{2^{\mu - 1} - 1}\Gamma\pars{\mu + 1}\zeta\pars{\mu}
\\[5mm]
\mc{J}_{2}'\pars{0} & = \bbx{\ds{-\gamma + \ln\pars{\pi \over 4}}}
\label{2}\tag{2}
\end{align}

\begin{align}
\mc{J}_{4}\pars{\mu} & =
2^{3 - \mu}\,\Gamma\pars{\mu + 1}\sum_{n = 0}^{\infty}\pars{-1}^{n}\,
{\pars{n + 3}\pars{n + 2}\pars{n + 1}/3! \over \pars{n + 2}^{\,\mu + 1}}
\\[5mm] & =
{1 \over 3}\,2^{2 - 2\mu}\Gamma\pars{\mu + 1}\bracks{%
4\zeta\pars{\mu - 2} - \zeta\pars{\mu} - 4\zeta\pars{\mu - 2,{3 \over 2}} +
\zeta\pars{\mu,{3 \over 2}}}
\\[5mm]
\mc{J}_{4}'\pars{0} & =
-\,{2 \over 3}\,\gamma - {2 \over 3}\,\ln\pars{4 \over \pi} +
{28 \over 3}\,\
\overbrace{\zeta'\pars{-2}}^{\ds{-\,{\zeta\pars{3} \over 4\pi^{2}}\phantom{-}}} =
\bbx{\ds{-\,{2 \over 3}\,\gamma - {2 \over 3}\,\ln\pars{4 \over \pi} -
{7 \over 3\pi^{2}}\,\zeta\pars{3}}}
\label{3}\tag{3}
\end{align}
$\ds{\zeta'\pars{-2}}$ is evaluated with
Riemann Functional Equation.

With \eqref{1}, \eqref{2} and \eqref{3}:
\begin{align}
\int_{0}^{\infty}{\tanh^{2}\pars{x} \over x^{2}}\,\dd x & =
-6\bracks{-\,{2 \over 3}\,\gamma - {2 \over 3}\,\ln\pars{4 \over \pi} -
{7 \over 3\pi^{2}}\,\zeta\pars{3}} +
4\bracks{-\gamma + \ln\pars{\pi \over 4}}
\\[5mm] & =
\bbx{\ds{14\zeta\pars{3} \over \pi^{2}}}
\end{align}
