<s>[INST] 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.
 [/INST] Firstly observe that, for $x$ real,
\begin{align}(1-x)^2-(1-x)+1&=(1-2x+x^2)-1+x+1\\
&=x^2-x+1
\end{align}
\begin{align}J&=\int_0^1 \frac{\ln(x^2-x+1)}{x(x-1)}\,dx\\
&=-\int_0^1 \frac{\ln(x^2-x+1)}{1-x}\,dx-\int_0^1 \frac{\ln(x^2-x+1)}{x}\,dx
\end{align}
In the first integral perform the change of variable $y=1-x$,
\begin{align}J&=-2\int_0^1 \frac{\ln(x^2-x+1)}{x}\,dx\\
&=-2\int_0^1 \frac{\ln\left(\frac{x^3+1}{x+1}\right)}{x}\,dx\\
&=2\int_0^1 \frac{\ln\left(x+1\right)}{x}\,dx-2\int_0^1 \frac{\ln\left(x^3+1\right)}{x}\,dx\\
&=2\int_0^1 \frac{\ln\left(x+1\right)}{x}\,dx-2\int_0^1 \frac{x^2\ln\left(x^3+1\right)}{x^3}\,dx\\
\end{align}
In the latter integral perform the change of variable $y=x^3$,
\begin{align}J&=2\int_0^1 \frac{\ln\left(x+1\right)}{x}\,dx-\frac{2}{3}\int_0^1 
\frac{\ln\left(x+1\right)}{x}\,dx\\
&=\frac{4}{3}\int_0^1 \frac{\ln\left(x+1\right)}{x}\,dx\\
&=\frac{4}{3}\int_0^1 \frac{\ln\left(1-x^2\right)}{x}\,dx-\frac{4}{3}\int_0^1 \frac{\ln\left(1-x\right)}{x}\,dx\\
&=\frac{4}{3}\int_0^1 \frac{x\ln\left(1-x^2\right)}{x^2}\,dx-\frac{4}{3}\int_0^1 \frac{\ln\left(1-x\right)}{x}\,dx
\end{align}
In the first integral perform the change of variable $y=x^2$,
\begin{align}J&=\frac{2}{3}\int_0^1 \frac{\ln\left(1-x\right)}{x}\,dx-\frac{4}{3}\int_0^1 \frac{\ln\left(1-x\right)}{x}\,dx\\
&=-\frac{2}{3}\int_0^1 \frac{\ln\left(1-x\right)}{x}\,dx\\
&=-\frac{2}{3}\Big[\ln x\ln(1-x)\Big]_0^1 -\frac{2}{3}\int_0^1 \frac{\ln x}{1-x}\,dx\\
&=-\frac{2}{3}\int_0^1 \frac{\ln x}{1-x}\,dx\\
&=\frac{2}{3}\zeta(2)\\
&=\frac{2}{3}\times \frac{\pi^2}{6}\\
&=\boxed{\frac{\pi^2}{9}}
\end{align}
</s>