Prove $\int_0^1\frac{\log(t^2-t+1)}{t^2-t}\mathrm dt=\frac{\pi^2}9$ I am in the middle of proving that 
$$\sum_{k\geq1}\frac1{k^2{2k\choose k}}=\frac{\pi^2}{18}$$
And I have reduced the series to 
$$\sum_{k\geq1}\frac1{k^2{2k\choose k}}=\frac12\int_0^1\frac{\log(t^2-t+1)}{t^2-t}\mathrm dt$$
But this integral is giving me issues. I broke up the integral
$$\int_0^1\frac{\log(t^2-t+1)}{t^2-t}\mathrm dt=\int_0^1\frac{\log(t^2-t+1)}{t-1}\mathrm dt-\int_0^1\frac{\log(t^2-t+1)}t\mathrm dt$$
I preformed the substitution $t-1=u$ on the first integral, then split it up:
$$\int_0^1\frac{\log(t^2-t+1)}{t-1}\mathrm dt=\int_{-1}^0\frac{\log(2u+i\sqrt3+1)}u\mathrm du+\int_{-1}^0\frac{\log(2u-i\sqrt3+1)}u\mathrm du-2\log2\int_{-1}^0\frac{\mathrm du}u$$
But the last term diverges, but I don't know what I did wrong. In any case, I would be surprised if there wasn't an easier way to go about this. Any suggestions? Thanks.

Another Approach is to employ Feynman's Trick:
Let
$$I(x) = \int_{0}^{1} \frac{\ln\left| x^2\left(t^2 - t\right) + 1\right|}{t^2 - t}\:dt$$
Note $I = I(1)$ and $I(0) = 0$
Thus
\begin{align}
I'(x) &= \int_{0}^{1} \frac{2x\left(t^2 - t\right)}{\left(x^2\left(t^2 - t\right) + 1\right)\left( t^2 - t\right)}\:dt = \frac{2}{x}\int_{0}^{1} \frac{1}{\left(t - \frac{1}{2}\right)^2 + \frac{4 - x^2}{4x^2}}\:dt\\
&= \frac{4}{x}\int_{0}^{\frac{1}{2}} \frac{1}{t^2 + \frac{4 - x^2}{4x^2}}\:dt = \frac{8}{\sqrt{4 - x^2}}\arctan\left(\frac{x}{\sqrt{4 -x^2}} \right) 
\end{align}
We now integrate to solve $I(x)$
$$I(x) = \int\frac{8}{\sqrt{4 - x^2}}\arctan\left(\frac{x}{\sqrt{4 -x^2}} \right)  \:dx = 4\left[\arctan\left( \frac{x}{\sqrt{4 - x^2}}\right) \right]^2 + C $$
Where $C$ is a constant of integration. As $I(0) = 0$ we find $C = 0$ and so:
$$I(x) = 4\left[\arctan\left( \frac{x}{\sqrt{4 - x^2}}\right) \right]^2$$
And finally 
$$ I = I(1) = 4\left[\arctan\left( \frac{1}{\sqrt{3}}\right) \right]^2 = \frac{\pi^2}{9}$$
