\[\Large\displaystyle \int_0^1\frac{\ln x\ln^2(1+x)}{x}\ \mathrm dx
\]
\(\Large\mathbf{Solution:}\)
\[\begin{align*}
\int_0^1\frac{\ln x\ln^2(1+x)}{x}\ \mathrm dx&=\frac{1}{2}\ln^2 x\ln^2(1+x)\Bigg|_0^1-\int_0^1\frac{\ln^2{x}\ln(1+x)}{1+x}\,\mathrm dx\\
&=-\int_0^1\sum_{k=1}^\infty (-1)^{k-1}H_{k}\,x^k\ln^2x\,\mathrm dx\\
&=-\sum_{k=1}^\infty (-1)^{k-1}H_{k}\int_0^1x^k\ln^2x\,\mathrm dx\\
&=-2\sum_{k=1}^\infty (-1)^{k-1}\frac{H_{k}}{(k+1)^3}\\
&=-2\sum_{k=1}^\infty (-1)^{k-1}\left[\frac{H_{k+1}}{(k+1)^3}-\frac{1}{(k+1)^4}\right]\\
&=2\sum_{k=1}^\infty (-1)^{k-1}\left[\frac{H_{k}}{k^3}-\frac{1}{k^4}\right]\\
&=\frac{11\pi^4}{180}-4\text{Li}_4 \left(\frac{1}{2} \right)-\frac{7\zeta(3)\ln 2}{2}+\frac{\pi^2\ln^2 2}{6}-\frac{\ln^4 2}{6}-2\eta(4)\\
&=\frac{11\pi^4}{180}-4\text{Li}_4 \left(\frac{1}{2} \right)-\frac{7\zeta(3)\ln 2}{2}+\frac{\pi^2\ln^2 2}{6}-\frac{\ln^4 2}{6}-\frac{7\zeta(4)}{4}\\
&=\large\boxed{\displaystyle \color{Blue} {\frac{\pi^4}{24}-4\text{Li}_4 \left(\frac{1}{2} \right)-\frac{7\zeta(3)\ln 2}{2}+\frac{\pi^2\ln^2 2}{6}-\frac{\ln^4 2}{6}}
}\end{align*}\]