TITLE: A integral from "irresistible integrals"
QUESTION [14 upvotes]: Evaluate :
$$\int_0^1\frac{x}{(1-x+x^2)^2}\ln \ln \frac{1}{x}\text{d}x$$
The answer on the book is 
$$-\frac{\gamma}{3}-\frac13\ln\frac{6\sqrt{3}}{\pi}+\frac{\pi\sqrt{3}}{27}\left(5\ln 2\pi-6\ln \Gamma\left(\frac16\right)\right)$$
Could anyone show a proof?
Refers to chapter 12 of the book pg237.

REPLY [10 votes]: If you're still interested, Ryan, here is a link to Adamchiks paper on the topic.
You can rewrite $\displaystyle \frac{x}{(x^{2}-x+1)^{2}}=\frac{x^{3}}{(x^{3}+1)^{2}}+\frac{2x^{2}}{(x^{3}+1)^{2}}+\frac{x}{(x^{3}+1)^{2}}$ and use the formulae he derives. 
One of which is $\displaystyle \int_{0}^{1}\frac{x^{n-1}}{(1+x^{n})^{2}}\log\left(\log(1/x)\right)dx=\frac{-1}{2n}\left(\gamma+\log(\frac{2n}{\pi})\right)$ 
Your integral is down near (43).