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.
 A: 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
$$
\begin{split}
-\int_0^1\frac{\log(t^2-t+1)}t\mathrm dt
&= -\int_0^1\frac{\log(1+t^3)-\log(1+t)}t\mathrm dt \\
&= \frac13\int_0^1\frac{\mathrm{Li}_1(-t^3)}{t^3}3t^2\mathrm dt
  - \int_0^1\frac{\mathrm{Li}_1(-t)}{t}\mathrm dt \\
&= \frac13 \int_0^1\frac{\mathrm{Li}_1(-t)}{t}\mathrm dt
  - \int_0^1\frac{\mathrm{Li}_1(-t)}{t}\mathrm dt \\
&= -\frac23 \int_0^1\frac{\mathrm{Li}_1(-t)}{t}\mathrm dt \\
&= -\frac23 \int_0^{-1}\frac{\mathrm{Li}_1(t)}{t}\mathrm dt \\
&= -\frac23 \mathrm{Li}_2(-1)
= -\frac23 \left(-\frac{\pi^2}{12}\right)
= \frac{\pi^2}{18}.
\end{split}
$$
A: 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: $\ \displaystyle{\ln(1+x)=\sum_{n=1}^\infty \frac{(-1)^{n-1}x^n}{n}}$
A: 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 
$$ I = I(1) = 4\left[\arctan\left( \frac{1}{\sqrt{3}}\right) \right]^2 = \frac{\pi^2}{9}$$
A: 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}
A: Let $I(a)=\int_0^1\frac{\ln(4\sin^2a \ (t^2-t)+1)}{t^2-t}dt$. Then
$$I’(a)= \int_0^1 \frac{2\cot a }{t^2-t+\frac14\csc^2a}dt
=8a
$$
and
$$\int_0^1\frac{\ln(t^2-t+1)}{t^2-t}dt=I(\frac\pi6)=\int_0^{\pi/6}8a\ da=\frac{\pi^2}9
$$
