Prove $\sum_{n=1}^\infty \frac{6}{n(3n-1)}=9 \ln 3-\sqrt{3} \pi$ This series represents the sum of reciprocal pentagonal numbers (multiplied by $3$). I got its value from Wolfram Alpha:
$$\sum_{n=1}^\infty \frac{6}{n(3n-1)}=9 \ln 3-\sqrt{3} \pi$$
My attempt: Turn the series into an integral:
$$\sum_{n=1}^\infty \frac{x^{3n}}{n}=-\ln(1-x^3)$$
$$\sum_{n=1}^\infty \frac{x^{3n-2}}{n}=-\frac{\ln(1-x^3)}{x^2}$$
$$\sum_{n=1}^\infty \frac{1}{n(3n-1)}=-\int_0^1 \frac{\ln(1-x^3)}{x^2} dx$$
I don't know how to solve this integral.
We can also transform the common term into two parts:
$$\frac{1}{n(3n-1)}=\frac{3}{3n-1}-\frac{1}{n}$$
But this gives us two diverging series, so it doesn't help.

We can do the indefinite integral. Start by the obvious guess 
$$\frac{d}{dx} \frac{-\log(1-x^3)}{x} = \frac{\log(1-x^3)}{x^2}+\frac{3x}{1-x^3}$$
Now we have to wipe out that second term.
$$
\frac{3x}{1-x^3}=\frac{1}{1-x}+\frac{x-1}{1+x+x^2} \\
\int\frac{1}{1-x}= -\log(1-x)
$$
leaving the term $\frac{x-1}{1+x+x^2}$ to deal with. A reasonable try would be subtracting off
$$
\frac{d}{dx} (\log{\sqrt{1+x+x^2}}) = \frac{x+\frac12}{1+x+x^2}
$$
And we are left with 
$$\frac12\int\frac1{1+x+x^2}dx$$
If you know that $$\frac{d}{dx} \tan^{-1}(a+b*x) = \frac{b}{1+(a+bx)^2}$$ and set that equal to $\zeta/(1+x+x^2)$ you can find $a, b$ and the multiplicative constant, to get
$$\frac{d}{dx}\left( \frac{\tan^{-1}(\frac{1+2x}{\sqrt{3}})} {\sqrt{3}} \right) =\frac12\int\frac1{1+x+x^2}dx$$
I may have a factor of $3$ error in this chain.  At any rate, you can now see that the $\sqrt{3}\pi$ in the answer comes from an arctan of $\frac{1+2\cdot 1}{\sqrt{3}}$, which is $\pi/3$.
Glue all these parts together and you have your indefinite integral.
