Evaluating a seemingly simple integral I'm trying to evaluate the following integral, which arised while attempting to find the sum of a series :
$$\int_{0}^{1} \frac{\ln(x)}{x-1} \ln(1+\sqrt{x})\text{d}x$$
I've tried unsuccessfully some substitutions, integration by parts, feynman integration... I'm not familiar with more advanced integration techniques like residues theorem etc. ,so maybe that's the way to go. 
Any hint, solution or partial solution would be nice !
 A: You can reduce your integral to two integrals which have already been solved on this site:
Let $x=t^2$, perform a partial fraction decomposition and integrate by parts in the second integral to find
\begin{align}
I &\equiv \int\limits_0^1 \frac{-\ln(x) \ln(1+\sqrt{x})}{1-x} \, \mathrm{d}x \\
&= 4 \int \limits_0^1 [-\ln(t) \ln(1+t)] \frac{t}{(1-t)(1+t)} \, \mathrm{d}t \\
&= 2 \int \limits_0^1 \frac{-\ln(t) \ln(1+t)}{1-t} \, \mathrm{d} t - 2 \int \limits_0^1 \frac{-\ln(t) \ln(1+t)}{1+t} \, \mathrm{d} t \\
&= 2 \int \limits_0^1 \frac{-\ln(t) \ln(1+t)}{1-t} \, \mathrm{d} t - \int \limits_0^1 \frac{\ln^2 (1+t)}{t} \, \mathrm{d} t \\
&\equiv 2 I_1 - I_2 \, .
\end{align}
In this question $I_2 = \frac{\zeta(3)}{4}$ is proved (FDP's answer does not use contour integration or the polylogarithm). $I_1 = \frac{\pi^2}{4} \ln(2) - \zeta(3)$ is derived here (the answer relies on contour integration though and it would be nice to have a simpler proof). Thus we obtain the result suggested in the comments:
$$ I = 2 \left(\frac{\pi^2}{4} \ln(2) - \zeta(3)\right) - \frac{\zeta(3)}{4} = \frac{\pi^2}{2} \ln(2) - \frac{9}{4} \zeta(3) \, . $$
