Evaluate $\int_0^1 \log \left( \frac{x^2+\sqrt{3}x+1}{x^2-\sqrt{3}x+1} \right) \frac{dx}{x} $ I'm trying to evaluate this integral but i'm having a lot of problems with the standards method.
$$\int_0^1 
\log
\left(
\frac{x^2+\sqrt{3}x+1}{x^2-\sqrt{3}x+1}
\right)
\frac{dx}{x}
$$
I've tried integration by parts, taylor expansion of the $\log$ function and some substitution.
Wolfram says that the results is $\frac{\pi^2}{3}$. I think that it divide the integral in four parts factorizing the argument of the logarithm with complex numbers, anyway it seems to take too long as approach.
Do you have any suggestions?
I've also noticed that the function is even so we could also write the integral as
$$
\frac{1}{2}
\int_{-1}^1 
\log
\left(
\frac{x^2+\sqrt{3}x+1}{x^2-\sqrt{3}x+1}
\right)
\frac{dx}{x}
$$
 A: Let us consider that for any $\alpha\in S^1$
$$ \int_{0}^{1}\frac{\log(1-\alpha x)}{x}\,dx = -\sum_{n\geq 1}\int_{0}^{1}\frac{\alpha^n x^{n}}{nx}\,dx=-\sum_{n\geq 1}\frac{\alpha^n}{n^2}=-\text{Li}_2(\alpha) $$
and by setting $\zeta=\exp\left(\frac{2\pi i}{12}\right)$ we have
$$ \int_{0}^{1}\log\left(\frac{1-\sqrt{3}x+x^2}{1+\sqrt{3}x+x^2}\right)\,\frac{dx}{x}=\int_{0}^{1}\left[\log(1-\zeta x)+\log(1-\zeta^{11} x)-\log(1-\zeta^5 x)-\log(1-\zeta^7 x)\right]\frac{dx}{x} $$
hence your integral equals
$$ \text{Li}_2(\zeta)+\text{Li}_2(\zeta^{11})-\text{Li}_2(\zeta^5)-\text{Li}_2(\zeta^7)=2\text{Re}\,\text{Li}_2(\zeta)-2\text{Re}\,\text{Li}_2(\zeta^5). $$
Last trick: $\text{Re}\,\text{Li}_2(e^{i\theta})$ is an elementary function, since it is the formal primitive of the sawtooth wave $\sum_{n\geq 1}\frac{\sin(nx)}{n}$ (piecewise-linear), hence a periodic and piecewise-parabolic function. Putting everything together, the outcome is simply $2\zeta(2)=\frac{\pi^2}{3}$.
A: Let $$I(\theta)=\int_0^1\frac{\ln(x^2+2x\cos\theta+1)}xdx,$$
differentiating it gives $$I'(\theta)=\int_0^1-\frac{2\sin\theta}{1+2x\cos\theta+x^2}dx\\
=-\theta$$
(since $\theta\in(-\pi,\pi)$)
The integral need to find is $$I\left(\frac16\pi\right)-I\left(\frac56\pi\right)=\int_{5/6\pi}^{1/6\pi}-\theta d\theta=\frac13\pi^2$$
