I like integrals

$\begin{aligned} I & = \int_{-2}^{-1} \frac{\ln (x^2-x)}{x} dx \quad \quad \small \color{#3D99F6}{\text{Let }u = - x} \\ & = \int_{2}^{1} \frac{\ln (u^2+u)}{u} du \\ & = \int_{2}^{1} \frac{\ln u + \ln (1+u)}{u} du \\ & = \color{#3D99F6}{\int_{2}^{1} \frac{\ln u}{u} du} + \int_{2}^{1} \frac{\color{#D61F06}{\ln (1+u)}}{u} du \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \small \color{#3D99F6}{\text{By integration by parts}} \\ & = \color{#3D99F6}{\left[\ln^2 u \right]_2^1 - \int_{2}^{1} \frac{\ln u}{u}du} + \int_{2}^{1} \frac{1}{u} \color{#D61F06}{\sum_{k=1} ^\infty(-1)^{k+1} \frac{u^k}{k}} du \quad \quad \space \small \color{#D61F06}{\text{Maclaurin series}} \\ & = \color{#3D99F6}{-\frac{\ln^2 2}{2}} + \int_{2}^{1} \sum_{k=1} ^\infty(-1)^{k+1} \frac{u^{k-1}}{k} du \\ & = -\frac{\ln^2 2}{2} + \sum_{k=1}^\infty(-1)^{k+1} \int_{2}^{1} \frac{u^{k-1}}{k} du \\ & = -\frac{\ln^2 2}{2} + \sum_{k=1}^\infty(-1)^{k+1} \left[\frac{u^k}{k^2}\right]_2^1 \\ & = -\frac{\ln^2 2}{2} + \sum_{k=1}^\infty(-1)^{k+1} \frac{1}{k^2} - \sum_{k=1}^\infty(-1)^{k+1} \frac{2^k}{k^2} \\ & = -\frac{\ln^2 2}{2} + \frac{1}{2} \sum_{k=1}^\infty \frac{1}{k^2} + \sum_{k=1}^\infty \frac{(-2)^k}{k^2} \\ & = -\frac{\ln^2 2}{2} + \frac{\pi^2}{12} + Li_2 (-2) \\ & = \frac{1}{12}\left(\pi^2-6\ln^2 2\right) + Li_2 (-2) \end{aligned}$

$\Rightarrow A+B+C+D+F+G = 12+2+6+2+2+2 = \boxed{26}$