Who's up to the challenge? 33

The key observation is to write the integral as $I=1/a\int_{0}^1 e^{ax} \ln x dx\\=\frac{1}{a}\sum_{n\ge 0}a^n \int_{0}^1 x^n \ln x dx$ where $a=\ln 2$ . Now, observe that $\int_{0}^1 x^n \ln x dx=-\int_{0}^{\infty} e^{-(n+1)x}x dx=-\frac{1}{(n+1)^2}$ Thus, the integral becomes, $-\frac{1}{a}\sum_{n\ge 0}\frac{a^n}{n!(n+1)^2}=-\frac{1}{a^2}\sum_{n\ge 1}\frac{a^n}{n!}$ Using a series representation of $\mathrm{li}(x)$ , we get $\mathrm{li}(e^x)=\gamma+\ln |x|+\sum_{n\ge 1}\frac{x^n}{nn!}$ Hence, the expression results into $\frac{\gamma+\ln(\ln 2)-\mathrm{li}(2)}{(\ln 2)^2}$ which results into the answer $\boxed{10}$ .