<s>[INST] Seeking alternative methods for $\int _0^1\frac{\ln \left(x^2-x+1\right)}{x\left(1-x\right)}\:dx$ I've solved this one by first tackling,
$$\int _0^{\infty }\frac{\ln \left(x^2-x+1\right)}{x\left(1-x\right)}\:dx$$
But i'd like to know other ways to solve it since the way i did it was a bit lengthy and not that straightforward.
 [/INST] $$\int_0^1\frac{\ln(x^2-x+1)}{x(1-x)}\ dx=\underbrace{\int_0^1\frac{\ln(x^2-x+1)}{1-x}\ dx}_{x\to 1-x}+\int_0^1\frac{\ln(x^2-x+1)}{x}\ dx$$
$$=2\int_0^1\frac{\ln(x^2-x+1)}{x}\ dx=2\underbrace{\int_0^1\frac{\ln(x^3+1)}{x}\ dx}_{x^3\to x}-2\int_0^1\frac{\ln(1+x)}{x}\ dx$$
$$=-\frac43\int_0^1\frac{\ln(1+x)}{x}\ dx=-\frac43\cdot\frac12\zeta(2)=-\frac{\pi^2}{9}$$

A different way to calculate the last integral is to use the identity
$$\sum_{n=1}^{\infty}\frac{x^n}{n}\cos(an)=-\frac12\ln(1-2x\cos(a)+x^2)$$
Set $a=\frac{\pi}{3}$ we get
$$\ln(1-x+x^2)=-2\sum_{n=1}^\infty \frac{x^n}{n}\cos(n\pi/3)$$
so
$$\int_0^1\frac{\ln(1-x+x^2)}{x}\ dx=-2\sum_{n=1}^\infty \frac{\cos(n\pi/3)}{n}\int_0^1 x^{n-1}\ dx$$
$$=2\sum_{n=1}^\infty\frac{\cos(n\pi/3)}{n^2}=-\frac{\pi^2}{18}$$
where the last result follows from integrating both sides of the common identity
$$\sum_{n=1}^\infty \frac{\sin(nx)}{n}=\frac{\pi-x}{2}$$
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