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To add one more solution link to this answer listing Feynman's trick exercises, I'm posting herein a calculation by said technique of $I_{12}:=\int_0^1\frac{\ln(1-x+x^2)dx}{x(1-x)}$, viz.$$\begin{align}I_{12}&=\int_0^{\pi/6}dt\frac{d}{dt}\int_0^1\frac{\ln(1-x(1-x)4\sin^2t)dx}{x(1-x)}\\&=-8\int_0^{\pi/6}dt\left[\sin t\cos t\int_0^1\frac{dx}{1-x(1-x)4\sin^2t}\right]\\&\stackrel{\ast}{=}-8\int_0^{\pi/6}tdt\\&=-\frac{\pi^2}{9},\end{align}$$where $\stackrel{\ast}{=}$ substitutes $x=\frac{1+\cot t\tan u}{2}$.

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