Evaluating $\int_0^{1/2} \log(1-x) \log(1-2x) \ dx$. While messing around with Wolfram Alpha, I discovered that 

$$\int_0^{\frac{1}2} \log(1-x) \log(1-2x) \ dx = 1 - \frac{\pi^2}{24} - \frac{\log(2)}2.$$

I've tried all sorts of standard tricks, but I cannot seem to prove it. Can someone prove this beautiful identity? 

With the substitution $u = 1-2x$, $x = (1+u)/2$, $dx = -du/2$, we get $$\begin{align*} \int_{x=0}^{1/2} \log (1-x) \log (1-2x) \, dx &= -\frac{1}{2} \int_{u=1}^0 \log \left(\frac{1+u}{2}\right) \log u \, du \\ &= \frac{1}{2} \int_{u=0}^1 \log u \,(\log (1+u) - \log 2) \, du \\
&= \frac{1}{2} \left( -\log 2 \int_{u=0}^1 \log u \, du + \int_{u=0}^1 \log u \, \log (1+u) \, du \right) \\ &= \frac{\log 2}{2} + \frac{1}{2} I,\end{align*}$$ where $$\begin{align*} I &= \int_{u=0}^1 \log u \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k} u^k \, du \\ &= \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k} \int_{u=0}^1  u^k \log u \, du \\
&= \sum_{k=1}^\infty \frac{(-1)^k}{k(k+1)^2}, \end{align*}$$ with the evaluation of the last step being accomplished by a trivial integration by parts.  Partial fraction decomposition gives $$\frac{1}{k(k+1)^2} = \frac{1}{k} - \frac{1}{k+1} - \frac{1}{(k+1)^2},$$ and using the fact that $$\sum_{k=1}^\infty \frac{(-1)^k}{k} = -\log 2,$$ and $$\sum_{k=1}^\infty \frac{1}{k^2} = \frac{\pi^2}{6},$$ it is relatively straightforward to obtain the desired result.
