Prove $\int_0^1\frac{\log(t^2-t+1)}{t^2-t}\mathrm dt=\frac{\pi^2}9$

Question

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.

Answer 1

Let's multiply by $1$ the integral found in @Federico's answer. $$\int_0^1\frac{\log(1-x+x^2)}x dx =\int_0^1\frac{\ln(1+x^3)-\ln(1+x)}{x}dx$$ $$\int_0^1\frac{\ln(1+x^3)}{x}dx\overset{x=t^{1/3}}=\frac13\int_0^1 \frac{\ln(1+t)}{t^{1/3}}\,t^{1/3-1}dt\overset{t=x}=\frac13\int_0^1\frac{\ln(1+x)}{x}dx$$ $$\sum_{n=1}^\infty \frac1{n^2{2n\choose n}} =\frac23 \int_0^1 \frac{\ln(1+x)}{x}dx=\frac23\sum_{n=1}^\infty \int_0^1\frac{(-1)^{n-1}x^{n-1}}{n}dx$$$$=\frac23\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^2}=\frac23\cdot\frac{\pi^2}{12}=\frac{\pi^2}{18}$$ Above I used: $\ \displaystyle{\ln(1+x)=\sum_{n=1}^\infty \frac{(-1)^{n-1}x^n}{n}}$