How to show that $ \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} \ln\left(\tan x+\tan\frac{\pi}{6}\right)\tan x\space dx=\frac{\zeta(2)}{6} $ I was trying to prove the well known result:
$$
\sum_{k=1}^\infty \frac{1}{\binom{2k}kk^2}=\frac{\zeta(2)}{3}
$$
and it came down to prove the following integral
$$
\int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} \ln\left(\tan x+\tan\frac{\pi}{6}\right)\tan x\space dx=\frac{\zeta(2)}{6}
$$
Can this integral be proven without using the above mentionend sum? I tried applying the angle sum and difference identities but I didn't get something helpful, so any help is highly appreciated!
Edit:
Can it be shown directly, in other words without knowing $\zeta(2)=\frac{\pi^2}{6}$, that
$$
\sum_{k=1}^\infty \frac{1}{\binom{2k}kk^2}=\frac{\zeta(2)}{3}
$$
 A: I am not sure if this helps to focus on the sum itself. If it does, then start with $\displaystyle f(x)=2\arcsin(\frac{x}{2})^2$. Write a Taylor's expansion for $\displaystyle f(x)$ (this is rather painful but fun):$$2\arcsin(\frac{x}{2})^2=\sum_{n=1}^{\infty}\frac{x^{2n}}{\binom{2n}n n^2}.$$ Now observe that $$2\arcsin(\frac{1}{2})^2=\sum_{n=1}^{\infty}\frac{1}{\binom{2n}n n^2},$$or that$$\sum_{n=1}^{\infty}\frac{1}{\binom{2n}n n^2}=2 \times (\frac{\pi}{6})^2.$$ 
A: Substitute $\tan x= t\tan\frac{\pi}{6}$
\begin{align}
&\int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} \ln\left(\tan x+\tan\frac{\pi}{6}\right)\tan x\space dx\\
=& \int_{-1}^1 \frac{t\ln (1+t)}{3+t^2}dt
= \int_{-1}^0 \frac{t\ln (1+t)}{3+t^2}\overset{t\to -t}{dt }+\int_{0}^1 \frac{t\ln (1+t)}{3+t^2}dt\\
=& \int_{0}^1 \frac{t\ln \frac{1+t}{1-t}}{3+t^2}dt
\overset{t\to \frac{1-t}{1+t}}=
-\frac12\int_0^1 \frac{(1-t)\ln t}{(1+t)(1+t+t^2)}dt \\
=& -\frac12\int_0^1 \ln t\>d\bigg(\ln \frac{(1-t^2)^2}{(1-t)(1-t^3)}\bigg)\\
\overset{ibp}=&
-\frac12\int_0^1 \frac{\ln(1-t)}tdt -\frac12\int_0^1 \frac{\ln(1-t^3)}t \overset{t^3\to t}{dt}+\int_0^1 \frac{\ln(1-t^2)}t\overset{t^2\to t}{dt}\\
=&\>-\frac16\int_0^1 \frac{\ln(1-t)}tdt= -\frac16(-\zeta(2))
=\frac{\zeta(2)}{6} 
\end{align}
where $\int_0^1 \frac{\ln(1-t)}tdt=-\sum_{k\ge 1} \frac1 k \int_0^1 t^{k-1}dt=-\sum_{k\ge 1} \frac1{k^2}=-\zeta(2)
$.
A: We use the notation $c = \dfrac{1}{\sqrt 3}$ for convenience. Substitute $t = \tan x$ to get
$$
I = \int_{-c}^c dt\frac{t \log(t+c)}{1+t^2}.
$$
There exists an antiderivative, namely 
$\newcommand{\lb}{\left(}$
$\newcommand{\rb}{\right)}$
$\newcommand{\lbb}{\left[}$
$\newcommand{\rbb}{\right]}$
$\newcommand{\li}{\operatorname{Li}_2}$
$$
I = \int dt\frac{t \log(t+c)}{1+t^2} = \Re \lbb \log \lb \frac{i-t}{i+c} \rb \log (t+c)+ \li\lb \frac{c+t}{c-i} \rb\rbb.
$$
Substituting the limits, we find that the antiderivative vanishes at the left endpoint and thus
$$
I = \underbrace{\Re \lbb \log \lb \frac{i-c}{i+c} \rb  \log (2c)\rbb}_0 + \Re \lbb \li\lb \frac{2c}{c-i} \rb\rbb \\= \Re \li\lb e^{ i \pi/3} \rb=\frac{1}{4}(\pi - \pi/3)^2-\frac{\pi^2}{12}=\frac{\pi^2}{36} = \frac {\zeta (2)}{ 6} .
$$
Here I used formula 5.16 from Lewin's book (polylogs and associated functions) to calculate the real part of the polylogarithm. The antiderivative can be straightforwardly calculated by writing 
$$
\frac{1}{1+t^2} = \frac 1 2 \lbb \frac{1}{1 - i t} + \frac{1}{1+ i t} \rbb = \Re \lbb \frac{1}{1 - i t} \rbb.
$$
