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.

My approach.
$$\int _0^{\infty }\frac{\ln \left(x^2-x+1\right)}{x\left(1-x\right)}\:dx$$
$$=\int _0^1\frac{\ln \left(x^2-x+1\right)}{x\left(1-x\right)}\:dx\:+\int _1^{\infty }\frac{\ln \left(x^2-x+1\right)}{x\left(1-x\right)}\:dx$$
Let $\displaystyle x=\frac{1}{t}$ for the $2$nd integral.
$$\int _0^1\frac{\ln \left(t^2-t+1\right)}{t\left(1-t\right)}\:dt\:+\int _0^1\frac{\ln \left(t^2-t+1\right)}{t-1}\:dt-2\int _0^1\frac{\ln \left(t\right)}{t-1}\:dt$$
$$=\int _0^1\left(\frac{1}{t\left(1-t\right)}+\frac{1}{t-1}\right)\ln \left(t^2-t+1\right)\:dt\:-2\sum _{k=0}^{\infty }\frac{1}{\left(k+1\right)^2}$$
$$=\int _0^1\frac{\ln \left(t^2-t+1\right)}{t}\:dt\:-\frac{\pi ^2}{3}$$
$$\int _0^1\frac{\ln \left(t^3+1\right)}{t}\:dt-\int _0^1\frac{\ln \left(t+1\right)}{t}\:dt-\frac{\pi ^2}{3}$$
$$\int _0^1\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k+1}}{k}t^{3k-1}\:dt-\int _0^1\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k+1}}{k}t^{k-1}\:dt-\frac{\pi ^2}{3}$$
$$\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k+1}}{3k^2}+\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k+2}}{k^2}-\frac{\pi ^2}{3}$$
$$\frac{\pi ^2}{36}-\frac{\pi ^2}{12}-\frac{\pi ^2}{3}=-\frac{7\pi ^2}{18}$$
So,
$$\boxed{\int _0^{\infty }\frac{\ln \left(x^2-x+1\right)}{x\left(1-x\right)}\:dx=-\frac{7\pi ^2}{18}}$$
In order to find the desired integral i used this previous expression.
$$\int _0^{\infty }\frac{\ln \left(x^2-x+1\right)}{x\left(1-x\right)}\:dx=\int _0^1\frac{\ln \left(t^2-t+1\right)}{t}\:dt\:-\frac{\pi ^2}{3}$$
And let $\displaystyle t=1-u$
$$-\frac{7\pi ^2}{18}=-\int _0^1\frac{\ln \left(u^2-u+1\right)}{u-1}\:du\:-\frac{\pi ^2}{3}$$
$$\boxed{\int _0^1\frac{\ln \left(u^2-u+1\right)}{u-1}\:du\:=\frac{\pi ^2}{18}}$$
Notice that on the $3$rd line all we have to do is to put in the result we just found and we'll be done.
$$-\frac{7\pi ^2}{18}=\int _0^1\frac{\ln \:\left(t^2-t+1\right)}{t\left(1-t\right)}\:dt\:+\frac{\pi ^2}{18}-\frac{\pi ^2}{3}$$
And finally.
$$\boxed{\int _0^1\frac{\ln \:\left(t^2-t+1\right)}{t\left(1-t\right)}\:dt\:=-\frac{\pi ^2}{9}}$$
