Integral $\int_0^1 \frac{\ln(1-\sqrt{3}t+t^2)}{t} dt$ $$I=\int_0^1 \frac{\ln(1-\sqrt{3}t+t^2)}{t} dt=-\frac{13}{72}\pi^2$$
Here is what I tried: $1-\sqrt{3}t+t^2=(z_1-t)(z_2-t)$ where $z_1=e^{\frac{\pi}{6}i}, z_2=e^{-\frac{\pi}{6}i}$
$$I=\int_0^1 \frac{\ln(z_1)+\ln(1-\frac{t}{z_1})}{t}+\frac{\ln(z_2)+\ln(1-\frac{t}{z_2})}{t} dt=\int_0^1 \frac{\ln(1-\frac{t}{z_1})}{t}+\frac{\ln(1-\frac{t}{z_2})}{t} dt$$
where $\ln(z_1)+\ln(z_2)=0$ and $\frac{1}{z_1}=z_2, \frac{1}{z_2}=z_1$
$$I=-\sum_{n=1}^\infty \frac{1}{n} \int_0^1 \frac{1}{t}\left(\frac{t}{z_1}\right)^n+\frac{1}{t}\left(\frac{t}{z_2}\right)^n dt = -\sum_{n=1}^\infty \frac{1}{n}\left( \frac{z_1^n+z_2^n}{n} \right)$$
where $z_1^n+z_2^n=2\cos(\frac{n\pi}{6})$
$$I=-2\sum_{n=1}^\infty \frac{\cos(\frac{n\pi}{6})}{n^2}$$
How to proceed next?

As @Frank W pointed out in comment...(although I saw the comment after I came up with the proof). Actually, I found the integral $I(a) = \pi a - \frac{a^2}{2} - \pi^2/3$ in the book "Table of Integrals, Series, and Products"
by I.S. Gradshteyn and I.M. Ryzhik. Then I came up with the proof.
For $a \in (0, \pi/2]$, let
$$I(a) := \int_0^1 \frac{\ln(1 - 2t \cos a + t^2)}{t}\,\mathrm{d} t.$$
We have
\begin{align*}
 I'(a) &= \int_0^1 \frac{2\sin a}{1 - 2t\cos a + t^2}\, \mathrm{d} t\\
 &= 2\arctan \frac{t - \cos a}{\sin a}\Big\vert_0^1 \\
 &= 2\arctan \frac{1 - \cos a}{\sin a}
 + 2\arctan\frac{\cos a}{\sin a}\\
 &= \pi - a
\end{align*}
We have $I(a) = \pi a - \frac{a^2}{2} + C$
on $(0, \pi/2]$ for some constant $C$.
Since $I(a)$ is continuous on $[0, \pi/2]$, taking limit as $a \to 0^{+}$, we get
$$C = \int_0^1 \frac{2\ln(1 - t)}{t}\,\mathrm{d} t = 
\int_0^1 (-2)\sum_{k=1}^\infty \frac{1}{k}t^{k-1}\,\mathrm{d} t = -2\sum_{k=1}^\infty \frac{1}{k^2} =  - \pi^2/3.$$
Thus, we have
$I(a) = \pi a - \frac{a^2}{2} - \pi^2/3$.
Thus, $I(\pi/6) = -\frac{13}{72}\pi^2$.
We are done.
