I need to evaluate the following integral with a high precision: $$ I=\int_{0}^{\infty}\left[% {\pi^{2} \over 6} - {\rm Li}_2\left({\rm e}^{-x}\right) -{\rm Li}_2\left({\rm e}^{-1/x}\right)\right]\,{{\rm d}x \over x}, $$ where ${\rm Li}_{2}$ denotes the dilogarithm $\displaystyle{% \left(~\mbox{note that}\ {\rm Li}_{2}\left(1\right) = {\pi^{2} \over 6}~\right)}$.
Unfortunately, a numerical integration in my CAS is only able to produce $3$ stable digits $I \approx 3.77\ldots$ that I do not even sure to be provably correct.
So, if only I were so lucky that a closed form existed for this integral, then, hopefully, it could be used to easily evaluate $I$ with a much higher precision. Could you suggest how to find a closed form ( if one exists )?
$$ I=\int^{\infty}_0 x^{-1}\sum^{\infty}_{k=1}\frac{1-e^{-kx}-e^{-kx^{-1}}}{k^2}dx\\ =\sum^{\infty}_{k=1}k^{-2}\int^{\infty}_0 \frac{1-e^{-kx}-e^{-kx^{-1}}}{x}dx\\ =2\sum^{\infty}_{k=1}k^{-2}(\gamma+\log k)\\ =2\gamma\zeta(2)-2\zeta'(2)\\ =2\zeta(2)(12\log A-\log2\pi). $$
Here $A$ is the Glaisher-Kinkelin constant.
Edit: $$\int^{\infty}_0 (1-e^{-kx}-e^{-kx^{-1}})\frac{dx}{x}\\ =\int^{1}_0 \frac{1-e^{-kx}}{x}dx-\int^{1}_0 \frac{e^{-kx^{-1}}}{x}dx+\int^{\infty}_1 \frac{1-e^{-kx^{-1}}}{x}dx-\int^{\infty}_1 \frac{e^{-kx}}{x}dx\\ =\int^{1}_0 \frac{1-e^{-kx}}{x}dx-\int^{\infty}_1 \frac{e^{-ky}}{y}dy+\int^{1}_0 \frac{1-e^{-ky}}{y}dy-\int^{\infty}_1 \frac{e^{-kx}}{x}dx\\ =2\int^{1}_0 \frac{1-e^{-kx}}{x}dx-2\int^{\infty}_1 \frac{e^{-kx}}{x}dx\\ =2\int^{k}_0 \frac{1-e^{-x}}{x}dx-2\int^{\infty}_k \frac{e^{-x}}{x}dx\\ =2(\operatorname{Ein}(k)-E_1(k))=2(\gamma+\log k). $$ Here $E_1$ and $\operatorname{Ein}$ are exponential integrals, see §6.2 of DLMF.