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

I start from $$ \int_0^1\frac{\log(t^2-t+1)}{t-1}\mathrm dt-\int_0^1\frac{\log(t^2-t+1)}t\mathrm dt . $$ In the first integral, substitute $t=1-u$. Then $$ \int_0^1\frac{\log(t^2-t+1)}{t-1}\mathrm dt =-\int_0^1\frac{\log(u^2-u+1)}{u}\mathrm du . $$

So you get $$ \sum_{k\geq1}\frac1{k^2{2k\choose k}} = -\int_0^1\frac{\log(t^2-t+1)}t\mathrm dt . $$

ADDENDUM

After some sleep, I managed to compute the integral with the help of polylogarithm. For $n\in\mathbb R$, define $$ \mathrm{Li}_n(z) = \sum_{k=1}^\infty \frac{z^k}{k^n}. $$

Some simple facts.

• $\mathrm{Li}_1(z)=-\log(1-z)$.
• $\mathrm{Li}_2(-1)=\sum_{k=1}^\infty \frac{(-1)^k}{k^2} = -\frac{\pi^2}{12}$.
• $z\frac{d}{dz} \mathrm{Li}_n(z) = \mathrm{Li}_{n-1}(z)$.
• $\mathrm{Li}_n(z) = \int_0^z \frac{\mathrm{Li}_{n-1}(s)}{s}\,ds$.

Then $$ \begin{split} -\int_0^1\frac{\log(t^2-t+1)}t\mathrm dt &= -\int_0^1\frac{\log(1+t^3)-\log(1+t)}t\mathrm dt \\ &= \frac13\int_0^1\frac{\mathrm{Li}_1(-t^3)}{t^3}3t^2\mathrm dt - \int_0^1\frac{\mathrm{Li}_1(-t)}{t}\mathrm dt \\ &= \frac13 \int_0^1\frac{\mathrm{Li}_1(-t)}{t}\mathrm dt - \int_0^1\frac{\mathrm{Li}_1(-t)}{t}\mathrm dt \\ &= -\frac23 \int_0^1\frac{\mathrm{Li}_1(-t)}{t}\mathrm dt \\ &= -\frac23 \int_0^{-1}\frac{\mathrm{Li}_1(t)}{t}\mathrm dt \\ &= -\frac23 \mathrm{Li}_2(-1) = -\frac23 \left(-\frac{\pi^2}{12}\right) = \frac{\pi^2}{18}. \end{split} $$

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: $\ \displaystyle{\ln(1+x)=\sum_{n=1}^\infty \frac{(-1)^{n-1}x^n}{n}}$

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

$$ I = I(1) = 4\left[\arctan\left( \frac{1}{\sqrt{3}}\right) \right]^2 = \frac{\pi^2}{9}$$