"I am Back" says Integration Part 5

If we define $\begin{array}{rcl} F(\alpha) & = & \displaystyle \int_0^\infty \frac{x^\alpha}{e^x-1}\,dx \; = \; \int_0^\infty x^\alpha e^{-x}(1 - e^{-x})^{-1}\,dx \\ & = & \displaystyle \sum_{n=0}^\infty \int_0^\infty x^{\alpha} e^{-(n+1)x}\,dx \; = \; \sum_{n=0}^\infty \frac{1}{(n+1)^{\alpha+1}} \int_0^\infty y^\alpha e^{-y}\,dy \\ & = & \zeta(\alpha+1) \Gamma(\alpha+1) \end{array}$ for $\alpha > -1$ , noting that the term-by-term integration is justified by the Monotone Convergence Theorem, then it is clear that $\begin{array}{rcl} \displaystyle I \; = \; \int_0^\infty \frac{x \ln x}{e^x-1}\,dx & = & \displaystyle F'(1) \; = \; \zeta'(2)\Gamma(2) + \zeta(2)\Gamma'(2) \\ & = & \zeta'(2) + \tfrac16\pi^2\Gamma'(2) \; = \; \tfrac16\pi^2\big[\gamma + \ln(2\pi) - 12\ln A\big] + \tfrac16\pi^2(1 - \gamma) \\ & = & \tfrac16\pi^2\big[ \gamma + \ln(2\pi) - 12\ln A - (\gamma-1)\big] \end{array}$ so that the answer is $2 + 6 + 2 + 12 + 1 = \boxed{23}$ . It is a bit strange that the answer does not cancel the $\gamma$ terms.