Who's up to the challenge? 70

We know , $\displaystyle -Li_{s}(-1) = \frac{1}{\Gamma(s)}\int_{0}^{\infty} \frac{x^{s-1}}{e^x+1}dx$

So if $\displaystyle G(s)=\int_{0}^{\infty} \frac{x^{s-1}}{e^x+1}dx$ then our target integral is $G'(2)$

$\displaystyle G(s) = -Li_{s}(-1)\Gamma(s) = \eta(s)\Gamma(s)$ , Since by definition of polygarithm we see $\displaystyle -Li_{s}(-1)=\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^s} = \eta(s)$

Since $\displaystyle\eta(s)=(1-2^{1-s})\zeta(s)$ & $\displaystyle \Gamma'(s)=\Gamma(s)\psi_{0}(s)=\Gamma(s)(H_{s-1}-\gamma)$ we have,

$\displaystyle G'(2) = \Gamma(2)\eta'(2) + \eta(2)\Gamma'(2) = \eta'(2) + \frac{\pi^2}{12}(1-\gamma)$

$\displaystyle \eta'(s) = \zeta'(s)(1-2^{1-s}) + 2^{1-s}\ln2\zeta(s)$

$\displaystyle \eta'(2) = \frac{\pi^2}{12}\gamma + \frac{\pi^2 \ln\pi}{12} + \frac{\pi^2 \ln2}{6} - \pi^2 \ln A$

So we have $\displaystyle G'(2) = \frac{\pi^2}{12} + \frac{\pi^2 \ln\pi}{12} + \frac{\pi^2 \ln2}{6} - \pi^2 \ln A$

So the answer is : $\boxed{12+2+2+12+2+2+6+2=40}$