Proof of $\int_{0}^{1}\dfrac{\log(x)\log(1 + 2x)}{x(1+x)} = -\dfrac{7}{6}\zeta(3)$ From quite sometime I've been struggling on proving
$$\int_{0}^{1}\dfrac{\log(x)\log(1 + 2x)}{x(1+x)}\,dx = -\dfrac{7}{6}\zeta(3)$$
I have tried partial fraction decomposition which produces a simple integral and a difficult one. I'm quite wondering if we can evaluate the integral without using the Linear combination of polylogarithms and logarithms. Any approach including complex analysis is most welcomed.
Any help would be appreciated. Thanks for reading.
 A: Note
\begin{align}
&\int_{0}^{1}\dfrac{\ln x\ln(1 + 2x)}{x(1+x)}dx
=\int_0^1 \int_0^2 \frac{\ln x}{(1+x)(1+t x)}dtdx\\
 =&\int_0^2 \frac{dt}{1-t}\int_0^1 \frac{\ln x}{1+x} -\frac{t\ln x}{1+ t x}\>dx
=\int_0^2 \frac{Li_2(-1)- Li_2(-t)}{1-t}dt\\
=& \int_0^1 \frac{Li_2(-1)- Li_2(-t)}{1-t}dt + \int_0^1 \frac{Li_2(-1-t)- Li_2(-1)}{t}dt\\
\overset{IBP}=&  \int_0^1\frac{\ln(1-t)\ln(1+t)}t dt+ \int_0^1\frac{\ln t\ln(2+t)}{1+t}dt
\end{align}
where $\int_0^1\frac{\ln t\ln(2+t)}{1+t}dt= -\frac{13}{24}\zeta(3)$ and
\begin{align}
\int_0^1 \frac{\ln(1-t)\ln(1+t)}{t}\,dx 
=&\frac14 \int_0^1 \overset{t^2\to t\to 1-t}{\frac{\ln^2(1-t^2)}t}dt
-\frac14 \int_0^1 \frac{\ln^2\overset{\to t}{\frac{1-t}{1+t}}}t dt\\
=& \left( \frac1{8} -\frac7{16}\right)\int_0^1 \frac{\ln^2 t}{1-t}dt =-\frac5{8}\zeta(3)
\end{align}
Thus
$$\int_{0}^{1}\frac{\ln x\ln(1 + 2x)}{x(1+x)}dx=-\frac58\zeta(3) -\frac{13}{24}\zeta(3)=-\frac76\zeta(3)$$
A: Assuming $0 \leq x\leq 1$ a CAS gives for the antiderivative
$$\text{Li}_3\left(1+\frac{1}{2
   x}\right)-\text{Li}_3\left(2+\frac{1}{x}\right)+\text{Li}_3(-2 x-1)+\text{Li}_3(-2
   x)+\text{Li}_3(-x)+\left(\text{Li}_2\left(2+\frac{1}{x}\right)-\text{Li}_2\left(1+
   \frac{1}{2 x}\right)\right) \log \left(\frac{1}{x}+2\right)-\text{Li}_2(-2 x-1)
   \log (2 x+1)-(\text{Li}_2(-2 x)+\text{Li}_2(-x)) \log (x)+\frac{1}{2} \log (2)
   \left(\log ^2(x)-\pi ^2\right)-\log (x) \log (2 (x+1)) \log (2 x+1)$$ I must confess that using the bounds I have a quite ugly expression
$$\text{Li}_3(-3)+\text{Li}_3(-2)+\text{Li}_3\left(\frac{3}{2}\right)-\text{Li}_3(3)-
   \text{Li}_2(-3) \log (3)+2 \text{Li}_2(3) \log (3)-\frac{\log ^3(3)}{2}+(\log (2)+i
   \pi ) \log ^2(3)+\frac{1}{6} \left(-\log ^3(2)+3 i \pi  \log ^2(2)+2\pi ^2 \log
   (2)\right)+\frac{1}{2} \left(\log \left(\frac{3}{2}\right)+i \pi \right)^2 \log
   (3)$$ which numerically is the rhs.
If we use partial fraction decomposition,
$$\int \frac{\log (x) \log (2 x+1)}{x (x+1)}\,dx=\int \frac{\log (x) \log (2 x+1)}{x }\,dx-\int \frac{\log (x) \log (2 x+1)}{ (x+1)}\,dx$$
we have for the simplest one
$$int \frac{\log (x) \log (2 x+1)}{x }\,dx=\text{Li}_3(-2 x)-\text{Li}_2(-2 x) \log (x)$$ but, as you wrote, the second is not pleasant
$$\int \frac{\log (x) \log (2 x+1)}{ (x+1)}\,dx=-\text{Li}_3\left(1+\frac{1}{2
   x}\right)+\text{Li}_3\left(2+\frac{1}{x}\right)-\text{Li}_3(-2
   x-1)-\text{Li}_3(-x)+\left(\text{Li}_2\left(1+\frac{1}{2
   x}\right)-\text{Li}_2\left(2+\frac{1}{x}\right)\right) \log
   \left(\frac{1}{x}+2\right)+\text{Li}_2(-2 x-1) \log (2 x+1)+\text{Li}_2(-x) \log
   (x)+\frac{1}{2} \log (2) \left(\pi ^2-\log ^2(x)\right)+\log (x) \log (2 (x+1))
   \log (2 x+1)$$ and still the same problem when using the bounds.
