Evaluate $\int_1^{\infty} {\frac{\ln{x}}{(x-1)(2x-1)}\,dx}$ Evaluate:
$$\int_1^{\infty} {\left(\frac{\ln{x}}{\left(x-1\right)\left(2x-1\right)}\,dx\right)}$$
It turns out that the value of the integral is exactly:
$$\frac{1}{12}\left(\pi^2+6\ln^2{2}\right)$$
as found by WolframAlpha, but Wolfram gives no indication of how it arrives at this curious result.
How would one solve the integral analytically? A first step might be the $u$-sub, $u=\frac{1}{x}$, which gives:
$$\int_1^{\infty} {\left(\frac{\ln{x}}{\left(x-1\right)\left(2x-1\right)}\,dx\right)}=-\int_0^1 {\left(\frac{\ln{x}}{\left(1-x\right)\left(2-x\right)}\,dx\right)}$$
Thanks!
 A: Start by performing the substitution $u=\frac 1x$, $\text dx=-\frac {1}{u^2}\text du$:
\begin{align}\int_1^\infty\left(\frac{\ln x}{(x-1)(2x-1)}\right)\text dx&=\int_1^0\left(\left(\frac{\ln\left(\frac 1u\right)}{\left(\frac 1u-1\right)\left(2\frac 1u-1\right)}\right)\times \left(-\frac {1}{u^2}\right)\right)\text du\\
&=-\int_0^1\left(\frac{-\ln\left(\frac 1u\right)}{(u-2)(u-1)}\right) du\\
\end{align}
We can now write the integral as follows, using partial fraction decomposition:
\begin{align}-\int_0^1\left(\frac{-\ln\left(\frac 1u\right)}{(u-2)(u-1)}\right) du&=-\int_0^1\left(-\ln\left(\frac 1u\right)\left(\frac{1}{(u-2)(u-1)}\right)\right)\text du\\
&=-\int_0^1\left(-\ln\left(\frac 1u\right)\left(\frac{1}{u-2}-\frac{1}{u-1}\right)\right)\text du\\
&=-\int_0^1\left(\ln\left(\frac 1u\right)\left(\frac{1}{u-1}-\frac{1}{u-2}\right)\right)\text du\end{align}
This now leaves us with:
\begin{align}-\int_0^1\left(\ln\left(\frac 1u\right)\left(\frac{1}{u-1}-\frac{1}{u-2}\right)\right)\text du&=-\left(\left(\int_0^1\ln\left(\frac 1u\right)\frac1{u-1}\right)\text du-\left(\int_0^1\ln\left(\frac 1u\right)\frac1{u-2}\right)\text du\right)\\
&=\left(\int_0^1\ln\left(\frac 1u\right)\frac1{u-2}\text du\right)-\left(\int_0^1\ln\left(\frac 1u\right)\frac1{u-1}\text du\right)\\
&=\left(\int_0^1-\ln(u)\frac1{u-2}\text du\right)-\left(\int_0^1-\ln(u)\frac1{u-1}\text du\right)\\
&=\left(-\int_0^1\frac{\ln u}{u-2}\text du\right)-\left(-\int_0^1\frac{\ln u}{u-1}\text du\right)\\
&=\left(\int_0^1\frac{\ln u}{u-1}\text du\right)-\left(\int_0^1\frac{\ln u}{u-2}\text du\right)\end{align}
Can you continue from here? At a glance, I would recommend trying integration by parts. 
A: This is related to the dilogarithm function $\text{Li}_2(x)$ and the special value $$\text{Li}_2(\frac{1}{2}) = \frac{\pi^2}{12}-\frac{\ln^2 2}{2}$$
The dilogarithm has integral representation:
$$\text{Li}_2(z) = \frac{\pi^2}{6}+\int_0^{1-z} \frac{\ln t}{1-t} dt$$
For your question, we have
$$\begin{aligned}
\int_0^1 \frac{\ln{x}}{\left(1-x\right)\left(2-x\right)}dx &= \int_0^1 \frac{\ln x}{1-x}dx - \int_0^1\frac{\ln x}{2-x}dx \\
&= \int_0^1 \frac{\ln x}{1-x}dx - 2\int_0^{1/2}\frac{\ln 2x}{2-2x}dx \\
&= \int_0^1 \frac{\ln x}{1-x}dx-(\ln 2) \int_0^{1/2}\frac{1}{1-x}dx-\int_0^{1/2}\frac{\ln x}{1-x}dx \\
&= -\frac{\pi^2}{6}-\ln^2 2 -[\text{Li}_2(\frac{1}{2})-\frac{\pi^2}{6}] \\
&= -\frac{\pi^2}{12}-\frac{\ln^2 2}{2}
\end{aligned}$$
