What is the value of the integral $\int_{0}^{\infty} \frac{x \ln (1+x^{2})}{\sinh (\pi x)} \, dx $? Mathematica obtains

$$\int_0^{\infty}\frac{x}{\sinh(\pi x)}\ln(1+x^2) \ \mathrm{d}x=-\frac{3}{4}+\frac{\ln 2}{6}+9\ln(A)-\frac{\ln \pi}{2}$$ where $A$ is the Glaisher-Kinkelin constant.

A numerical approximation of the integral strongly suggests that this is incorrect. What is the correct value?
 A: If I may add something that may help.
It would appear this integral can be tied to the integrals of log-Gamma or Barnes G in some form or another.
If we write the $\displaystyle csch(\pi x)$ in terms of its exponential, it is then very similar to the known result that can be found by using Hermite's formula for the Hurwitz zeta.
Hermite's formula for $\zeta(s,1)$ is:
$\displaystyle \zeta(s,1)=1/2+\frac{1}{s-1}+2\int_{0}^{\infty}\frac{\sin(s\cdot \tan^{-1}(x))}{(x^{2}+1)^{s/2}(e^{2\pi x}-1)}dx$.
Which can be differentiated w.r.t s, leading to:
$\displaystyle \int_{0}^{\infty}\frac{x\log(x^2+1)}{e^{2\pi x}-1}dx=1/2\log(2\pi)-2/3-\log(A)$.
The integral in question is also equal to the series:
$\displaystyle \sum_{n=2}^{\infty}(-1)^{n+1}\log\left(1-\frac{1}{n}\right)+\sum_{n=2}^{\infty}(-1)^{n+1}n\cdot \log\left(1+\frac{1}{n-1}\right)$
I will look into it later when I find more time. But, if anyone can use these to derive the solution, please feel free. 
A: Cody's answer gave me the idea to look at $\displaystyle \int_{0}^{\infty} \frac{\sin ( s \arctan t)}{(1+t^{2})^{s/2} \sinh \pi t} \, dt $.
First add the restriction $ \text{Re}(s) >1$.
Then 
$$ \begin{align} &\int_{0}^{\infty} \frac{\sin ( s \arctan t)}{(1+t^{2})^{s/2} \sinh \pi t} \, dt \\ &= \frac{1}{2}\int_{-\infty}^{\infty} \frac{\sin(s \arctan t)}{(1+t^{2})^{s/2} \sinh \pi t} \, dt \\&= \frac{1}{2} \int_{-\infty}^{\infty} \text{Im} \ \frac{1}{(1-it)^{s} \sinh \pi t} \, dt \\ &= \frac{1}{2} \, \text{Im} \, \text{PV} \int_{-\infty}^{\infty} \frac{1}{(1-it)^{s} \sinh \pi t} \, dt \\ &= \frac{1}{2} \, \text{Im} \left(\pi i \ \text{Res}\left[\frac{1}{(1-iz)^{s} \sinh \pi z},0\right] + 2 \pi i \sum_{n=1}^{\infty} \text{Res} \left[\frac{1}{(1-iz)^{s} \sinh \pi z},in \right] \right) \\ &= \frac{1}{2} \text{Im}  \left(\pi i \left(\frac{1}{\pi} \right) + 2 \pi i \sum_{n=1}^{\infty} \frac{(-1)^{n}}{\pi (1+n)^{s}} \right) \\ &= \frac{1}{2} \, \text{Im} \Big(i + 2i \big( (1-2^{1-s}) \zeta(s)-1\big) \Big) \\& = (1-2^{1-s})\zeta(s) - \frac{1}{2}. \end{align}$$
By analytic continuation, the result is valid for all complex values of $s$.
Differentiating under the integral sign and letting $s=-1$ we get
$$ \begin{align} \frac{1}{2} \int_{0}^{\infty} \frac{t \log(1+t^{2})}{\sinh \pi t} \, dt + \int_{0}^{\infty} \frac{\arctan t}{\sinh \pi t} \, dt  &=  2^{1-s} \log 2 \ \zeta(s) + (1-2^{1-s}) \zeta'(s) \Bigg|_{s=-1} \\ &= 4 \log (2) \left(-\frac{1}{12} \right) - 3 \zeta'(-1) \\ &= - \frac{\log 2}{3} - 3 \zeta'(-1) . \end{align}$$
So we need to evaluate $ \displaystyle \int_{0}^{\infty} \frac{\arctan t}{\sinh \pi t} \, dt $.
Let $ \displaystyle I(z) = \int_{0}^{\infty} \frac{\arctan \frac{x}{z}}{\sinh \pi x} \, dx $.
Then 
$$ \begin{align} I(z) &= \int_{0}^{\infty} \frac{1}{\sinh \pi x} \int_{0}^{\infty} \frac{\sin tx}{t} e^{-zt} \, dt \, dx \\ &= \int_{0}^{\infty} \frac{e^{-zt}}{t} \int_{0}^{\infty}  \frac{\sin t x}{\sinh \pi x} \, dx \, dt \\ &= \frac{1}{2} \int_{0}^{\infty} \frac{e^{-zt}}{t} \tanh \left( \frac{t}{2} \right) \, dt . \end{align} $$ 
Now differentiate under the integral sign.
$$ \begin{align} I'(z) &= - \frac{1}{2} \int_{0}^{\infty} \tanh \left(\frac{t}{2} \right) e^{-zt} \, dt \\ &= - \frac{1}{2} \int_{0}^{\infty} \frac{1-e^{-t}}{1+e^{-t}} e^{-zt} \, dt \\ &= - \frac{1}{2} \int_{0}^{\infty} \left( -1 + 2 \sum_{n=0}^{\infty} (-1)^{n} e^{-tn}\right) e^{-zt} \, dt \\ &= \frac{1}{2z} - \sum_{n=0}^{\infty} (-1)^{n} \frac{1}{z+n} \\ &=  \frac{1}{2z} - \frac{1}{2} \psi \left(\frac{z+1}{2} \right) + \frac{1}{2} \psi \left( \frac{z}{2} \right) \tag{1} \end{align}$$
Integrating back and using Stirling's formula to determine the value of the constant of integration, we find
$$ I(z) = \frac{\ln z}{2} -  \log \Gamma \left(\frac{z+1}{2} \right) + \log \Gamma \left(\frac{z}{2} \right) -  \frac{\log 2}{2} .$$
So 
$$ \int_{0}^{\infty} \frac{\arctan x}{\sinh \pi x} \, dx = I(1) = \frac{\log \pi}{2} - \frac{\log 2}{2} .$$
Combining this result with the first result we have
$$ \begin{align} \int_{0}^{\infty} \frac{x \log(1+x^{2})}{\sinh \pi x} \, dx  &= 2 \left(-\frac{\log 2}{3} - 3 \zeta'(-1) - \frac{\log \pi}{2} + \frac{\log 2}{2} \right) \\&=\frac{\log 2}{3} - \log \pi - 6 \zeta'(-1) \\ &= \frac{\log 2}{3} - \log \pi - \frac{1}{2} + 6 \log A \\ &\approx 0.0788460364 . \end{align}$$
$ $
$(1)$ http://mathworld.wolfram.com/DigammaFunction.html (5)
A: Noting the tabulated cosine transforms of $\ln(1+1/x^2)$ and $x/\sinh(\pi x)$, by the Parseval  relation for this transform
$$\int_0^{\infty}\frac{x \ln(1+1/x^2)}{\sinh(\pi x)}dx=\int_0^{\infty}\frac{dy}{y}\frac{1+\sinh y -\cosh y}{1+\cosh y}$$ which has the known value $12\ln A-\ln \pi -1-\frac{1}{3}\ln 2$.  It is also known that 
$$\int_0^{\infty}\frac{x \ln x^2}{\sinh(\pi x)}dx=\frac{1}{2}+\frac{2}{3}\ln 2-6\ln A =.$$ By combining these results one gets the desired  value. This points to a disturbing bug in Mathematica.
