Evaluation of $\int_0^1 \frac{\ln(1-x+x^2)}{x(1-x)}dx$

Question

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 $x=\frac{1+\cot t\tan u}{2}$.

Answer 1

Solution without so-called "Feynman's trick". \begin{align}J&=\int_0^1\frac{\ln(1-x+x^2)}{x(1-x)}dx\\ &=\int_0^1\frac{\ln(1-x+x^2)}{x}dx+\underbrace{\int_0^1\frac{\ln(1-x+x^2)}{1-x}dx}_{u=1-x}\\ &=2\int_0^1\frac{\ln(1-x+x^2)}{x}dx\\ &=2\int_0^1\frac{\ln\left(\frac{1+u^3}{1+u}\right)}{u}du\\ &=2\underbrace{\int_0^1\frac{\ln\left(1+u^3\right)}{u}du}_{w=u^3}-2\int_0^1\frac{\ln\left(1+u\right)}{u}du\\ &=\frac23\int_0^1\frac{\ln\left(\frac{1-w^2}{1-w}\right)}{w}dw-2\int_0^1\frac{\ln\left(\frac{1-u^2}{1-u}\right)}{u}du\\ &=\frac43\int_0^1\frac{\ln(1-w)}{w}dw-\frac43\underbrace{\int_0^1\frac{\ln(1-w^2)}{w}dw}_{z=w^2}\\ &=\frac23\int_0^1\frac{\ln(1-w)}{w}dw=\boxed{-\frac{\pi^2}{9}} \end{align} NB: I assume, \begin{align}\int_0^1\frac{\ln(1-x)}{x}dx=-\frac{\pi^2}6\end{align}