Too many parts, part 2 [pun intended] (My fifteenth integral problem)

After determined integration by parts, we have $\int x^2 e^{ax} \ln x\,dx \; = \; \big(\tfrac1a x^2 - \tfrac2{a^2} x\big)e^{ax}\ln x + \tfrac{2}{a^2}(e^{ax} - 1)\ln x - \big(\tfrac{1}{a^2} - \tfrac{3}{a^3}\big)e^{ax} - \tfrac{2}{a^3} \int \frac{e^{ax}-1}{x}\,dx$ and hence $\begin{array}{rcl} \displaystyle \int_0^1 x^2 e^{ax} \ln x\,dx & = & \displaystyle -\big(\tfrac{1}{a^2} - \tfrac{3}{a^3}\big)e^a - \tfrac{3}{a^3} - \tfrac{2}{a^3}\int_0^1 \frac{e^{ax}-1}{x}\,dx \\ & = & \displaystyle -\big(\tfrac{1}{a^2} - \tfrac{3}{a^3}\big)e^a - \tfrac{3}{a^3} - \tfrac{2}{a^3}\int_0^a \frac{e^x-1}{x}\,dx \\ & = & \displaystyle -\big(\tfrac{1}{a^2} - \tfrac{3}{a^3}\big)e^a - \tfrac{3}{a^3} - \tfrac{2}{a^3}\big[\mathrm{li}(e^a) - \ln a - \gamma\big] \\ & = & \displaystyle \frac{-2\mathrm{li}(e^a) + 3(e^a-1) + 2\gamma + \ln a^2 - ae^a}{a^3} \end{array}$ for any $a > 0$ . Putting $a = \ln2$ , we obtain $\begin{array}{rcl} \displaystyle \int_0^1 x^2 2^x \ln x\,dx & = &\displaystyle \frac{-2\mathrm{li}(2) + 3 + 2\gamma + \ln\big(\frac14(\ln2)^2\big)}{(\ln 2)^3} \end{array}$ making the answer $2 + 2 + 3 + 2 + 2 + 2 + 4 + 3 + 2 \,=\, \boxed{22}$ .

Alternatively, with less integration by parts, we could evaluate $\int_0^1 e^{ax} \ln x\,dx \; = \; \frac{\gamma + \ln a - \mathrm{li}(e^a)}{a}$ and differentiate this twice with respect to $a$ .