Closed form $\int_0^\infty\left(\frac{\tanh x}{x^2}-\frac{1}{xe^{2x}}\right)dx=12\log A-\frac{4}{3}\log 2$ Evaluate
$$\int_0^\infty\left(\frac{\tanh x}{x^2}-\frac{1}{xe^{2x}}\right)dx$$
I haven't been able to find references to the indefinite integral of the $\tanh$ term except for some similar forms that had solutions. see here and here.
Edit:
Following Random Variable's result we have the form
$$12\log A-\frac{4}{3}\log 2$$
 A: Let $$I(z) =\frac{z}{a}\int_0^\infty \left(\frac{1}{2}-\frac{z}{at}+\frac{1}{e^{at/z}-1}\right) \frac{1-e^{-at}}{t^2} dt$$
Then Binet's first formula says
$$I(z) = \int_0^z \left[\ln\Gamma(x) - (z-\frac{1}{2})\ln z + z - \frac{\ln(2\pi)}{2}\right] dx $$

Letting $a=2,z=1/2$ gives
$$4I(\frac{1}{2}) =\int_0^\infty \left(\frac{1}{2}-\frac{1}{4t}+\frac{1}{e^{4t}-1}\right) \frac{1-e^{-2t}}{t^2} dt$$
and $a=4,z=1$ gives
$$4I(1) =\int_0^\infty \left(\frac{1}{2}-\frac{1}{4t}+\frac{1}{e^{4t}-1}\right) \frac{1-e^{-4t}}{t^2} dt$$
Some algebraic manipulation yields
$$4I(1)-4I(\frac{1}{2}) = \underbrace{\int_0^\infty \left[\frac{1}{2t^2}-\frac{e^{-2t}}{2t} - (\frac{1}{2}-\frac{1}{4t})\left(\frac{e^{-2t}-e^{-4t}}{t^2}\right)\right] dt}_{J} - \frac{I}{2}$$
With $I$ your desired integral. Surprisingly, $J$ has elementary primitive (even it were not, we still have systematic way to crash it), with value $5/8$. 

Hence it remains to evalute $$\int_0^1 \ln \Gamma(x) dx \quad \quad \int_0^{1/2} \ln \Gamma(x) dx$$
The former is just $\ln(2\pi)/2$, for the latter, we can use the integral representation of Barnes G function:
$$\int_0^z \ln \Gamma(x) dx = \frac{z(1-z)}{2}+\frac{z}{2}\ln(2\pi) + z \ln\Gamma(z) - \log G(1+z)$$
and the special value $$\ln G(\frac{3}{2}) = -\frac{3}{2}\ln A + \frac{\ln \pi}{4}+\frac{1}{8}+\frac{\ln 2}{24}$$
with $A$ being the Glaisher-Kinkelin constant. 
Alternatively, use Fourier expansion of $\ln \Gamma(x)$, integrate termwise, and remember the relation between $A$ and $\zeta'(2)$ also gives the value of the integral $$\int_0^{1/2} \ln\Gamma(x)dx = \frac{3}{2}\ln A + \frac{5}{24}\ln 2 + \frac{\ln \pi}{4}$$
A: Let $$I(a,b) = \int_{0}^{\infty} \left(\frac{\tanh (x)}{x} -e^{-bx} \right)\frac{e^{-ax}}{x} \, dx, $$ where $a, b >0$.
I will show that $$I(a,b) = 8 \psi^{(-2)} \left(\frac{a+2}{4} \right)- 8 \psi^{(-2)}\left(\frac{a}{4} \right)-a \log(a)+a + 2 a \log (2)+ \log(a+b)- 2 \log(4 \pi),$$ where $\psi^{(-2)}(x)$ is the polygamma function of order $-2$ defined by the integral  $$ \psi^{(-2)}(x) = \int_{0}^{x} \log \Gamma (t) \, dt.$$
If we let $a \to 0^{+}$, we get $$\int_{0}^{\infty} \left(\frac{\tanh (x)}{x} -e^{-bx} \right)\frac{dx}{x}  =  8 \psi^{(-2)} \left(\frac{1}{2} \right)+ \log(b) - 2 \log(4\pi). $$
As explained in pisco's answer, $\psi^{(-2)} \left(\frac{1}{2} \right)$ can be expressed in terms of the Glaisher-Kinkelin constant.

Differentiating $I(a,b)$ under the integral sign with respect to $a$ (which is permissible for $a \ge c$, where $c$ is some positive value),  we get $$ \frac{\partial}{\partial a}I(a,b) = -\int_{0}^{\infty}\left(\frac{\tanh (x)}{x}-e^{-bx} \right) e^{-ax} \, dx = -\int_{0}^{\infty} \frac{\tanh (x)}{x} \, e^{-ax} \, dx + \frac{1}{a+b}.$$
And using a property of the Laplace transform, we get
$$\begin{align}\int_{0}^{\infty} \frac{\tanh (x)}{x} \, e^{-ax} \, dx &= \int_{a}^{\infty} \int_{0}^{\infty} \tanh (x) e^{-px} \, dx \, dp \\ &=\int_{a}^{\infty} \int_{0}^{\infty} \frac{1-e^{-2x}}{1+e^{-2x}} \, e^{-px} \, dx \, dp \\ &= \int_{a}^{\infty} \int_{0}^{\infty} \left(2\sum_{n=0}^{\infty} (-1)^{n} e^{-2nx} -1\right) e^{-px} \, dx \, dp \\ &= \int_{a}^{\infty} \left( 2 \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n+p}- \frac{1}{p}\right) \, dp \\ &= \int_{a}^{\infty} \left(\frac{2}{p} \sum_{n=0}^{\infty} \frac{(-1)^{n}}{\frac{2}{p}n +1} - \frac{1}{p}\right) \, dp \\ &= \int_{a}^{\infty} \left(\frac{1}{2} \left(\psi \left(\frac{p+2}{4} \right)- \psi\left(\frac{p}{4} \right)\right)- \frac{1}{p} \right) \, dp \tag{1} \\ &= -2 \log (2) -2\log \Gamma \left(\frac{a+2}{4} \right) +2 \log \Gamma \left(\frac{a}{4} \right) + \log (a). \tag{2} \end{align}$$
Therefore,  $$I(a,b) =2a \log(2) + 8 \psi^{(-2)} \left(\frac{a+2}{4} \right)- 8 \psi^{(-2)} \left(\frac{a}{4} \right)- a \log(a) +a + \log(a+b) +C. $$
To determine the integration constant $C$, we can take the limit on both sides of the above equation as $a \to +\infty$.
Since $\psi^{(-2)}(x) $ can be expressed in terms of the Barnes G-function, we can use the asymptotic expansion of the Barnes G-function.
(Term-by-term integration of Stirling's formula for the log gamma function does lead to the same expansion but with an unknown constant.)
The integration constant $C$ turns out to be $-2 \log(4 \pi)$.

$(1)$ https://mathworld.wolfram.com/DigammaFunction.html (6)
$(2)$ Using Stirling's formula, we have
$$\log \Gamma\left(\frac{a+2}{4} \right) \sim \left(\frac{a+2}{4} - \frac{1}{2} \right) \log \left(\frac{a+2}{4} \right) - \frac{a+2}{4} + \frac{1}{2} \log(2 \pi) + \mathcal{O}\left(\frac{1}{a} \right),  $$ where
$$\log \left(\frac{a+2}{4} \right)= \log \left(\frac{a}{4} \right) +  \log \left(1+ \frac{2}{a} \right) \sim \log \left(\frac{a}{4} \right) + \frac{2}{a} + \mathcal{O}\left(\frac{1}{a^{2}} \right).$$
