Start with integration by parts (IBP) by setting $u=\ln^3(1+x)$ and $dv=\dfrac{\ln x}{x}\ dx$ yields
\begin{align}
I&=-\frac32\int_0^1\frac{\ln^2(1+x)\ln^2 x}{1+x}\ dx\\
&=-\frac32\int_1^2\frac{\ln^2x\ln^2 (x-1)}{x}\ dx\quad\Rightarrow\quad\color{red}{x\mapsto1+x}\\
&=-\frac32\int_{\large\frac12}^1\left[\frac{\ln^2x\ln^2 (1-x)}{x}-\frac{2\ln^3x\ln(1-x)}{x}+\frac{\ln^4x}{x}\right]\ dx\quad\Rightarrow\quad\color{red}{x\mapsto\frac1x}\\
&=-\frac32\int_{\large\frac12}^1\frac{\ln^2x\ln^2 (1-x)}{x}\ dx+3\int_{\large\frac12}^1\frac{\ln^3x\ln(1-x)}{x}\ dx-\left.\frac3{10}\ln^5x\right|_{\large\frac12}^1\\
&=-\frac32\color{red}{\int_{\large\frac12}^1\frac{\ln^2x\ln^2 (1-x)}{x}\ dx}+3\int_{\large\frac12}^1\frac{\ln^3x\ln(1-x)}{x}\ dx-\frac3{10}\ln^52.
\end{align}
Applying IBP again to evaluate the red integral by setting $u=\ln^2(1-x)$ and $dv=\dfrac{\ln^2 x}{x}\ dx$ yields
\begin{align}
\color{red}{\int_{\large\frac12}^1\frac{\ln^2x\ln^2 (1-x)}{x}\ dx}&=\frac13\ln^52+\frac23\color{blue}{\int_{\large\frac12}^1\frac{\ln^3x\ln (1-x)}{1-x}\ dx}.
\end{align}
For the simplicity, let
$$
\color{blue}{\mathbf{H}_{m}^{(k)}(x)}=\sum_{n=1}^\infty \frac{H_{n}^{(k)}x^n}{n^m}\qquad\Rightarrow\qquad\color{blue}{\mathbf{H}(x)}=\sum_{n=1}^\infty H_{n}x^n,
$$
Introduce a generating function for the generalized harmonic numbers for $|x|<1$
$$
\color{blue}{\mathbf{H}^{(k)}(x)}=\sum_{n=1}^\infty H_{n}^{(k)}x^n=\frac{\operatorname{Li}_k(x)}{1-x}\qquad\Rightarrow\qquad\color{blue}{\mathbf{H}(x)}=-\frac{\ln(1-x)}{1-x}
$$
and the following identity
$$
H_{n+1}^{(k)}-H_{n}^{(k)}=\frac1{(n+1)^k}\qquad\Rightarrow\qquad H_{n+1}-H_{n}=\frac1{n+1}
$$
Let us integrating the indefinite form of the blue integral.
\begin{align}
\color{blue}{\int\frac{\ln^3x\ln (1-x)}{1-x}\ dx}=&-\int\sum_{n=1}^\infty H_nx^n\ln^3x\ dx\\
=&-\sum_{n=1}^\infty H_n\int x^n\ln^3x\ dx\\
=&-\sum_{n=1}^\infty H_n\frac{\partial^3}{\partial n^3}\left[\int x^n\ dx\right]\\
=&-\sum_{n=1}^\infty H_n\frac{\partial^3}{\partial n^3}\left[\frac{x^{n+1}}{n+1}\right]\\
=&-\sum_{n=1}^\infty H_n\left[\frac{x^{n+1}\ln^3x}{n+1}-\frac{3x^{n+1}\ln^2x}{(n+1)^2}+\frac{6x^{n+1}\ln x}{(n+1)^3}-\frac{6x^{n+1}}{(n+1)^4}\right]\\
=&-\ln^3x\sum_{n=1}^\infty \frac{H_{n+1}x^{n+1}}{n+1}+\ln^3x\sum_{n=1}^\infty \frac{x^{n+1}}{(n+1)^2}+3\ln^2x\sum_{n=1}^\infty \frac{H_{n+1}x^{n+1}}{(n+1)^2}\\&-3\ln^2x\sum_{n=1}^\infty \frac{x^{n+1}}{(n+1)^3}-6\ln x\sum_{n=1}^\infty \frac{H_{n+1}x^{n+1}}{(n+1)^3}+6\ln x\sum_{n=1}^\infty \frac{x^{n+1}}{(n+1)^4}\\&+6\sum_{n=1}^\infty \frac{H_{n+1}x^{n+1}}{(n+1)^4}-6\sum_{n=1}^\infty \frac{x^{n+1}}{(n+1)^5}\\
=&\ -\sum_{n=1}^\infty\left[\frac{H_nx^{n}\ln^3x}{n}-\frac{x^{n}\ln^3x}{n^2}-\frac{3H_nx^{n}\ln^2x}{n^2}+\frac{3x^{n}\ln^2x}{n^3}\right.\\& \left.\ +\frac{6H_nx^{n}\ln x}{n^3}-\frac{6x^{n}\ln x}{n^4}-\frac{6H_nx^{n}}{n^4}+\frac{6x^{n}}{n^5}\right]\\
=&\ -\color{blue}{\mathbf{H}_{1}(x)}\ln^3x+\operatorname{Li}_2(x)\ln^3x+3\color{blue}{\mathbf{H}_{2}(x)}\ln^2x-3\operatorname{Li}_3(x)\ln^2x\\&\ -6\color{blue}{\mathbf{H}_{3}(x)}\ln x+6\operatorname{Li}_4(x)\ln x+6\color{blue}{\mathbf{H}_{4}(x)}-6\operatorname{Li}_5(x).
\end{align}
Therefore
\begin{align}
\color{blue}{\int_{\Large\frac12}^1\frac{\ln^3x\ln (1-x)}{1-x}\ dx}
=&\ 6\color{blue}{\mathbf{H}_{4}(1)}-6\operatorname{Li}_5(1)-\left[\color{blue}{\mathbf{H}_{1}\left(\frac12\right)}\ln^32-\operatorname{Li}_2\left(\frac12\right)\ln^32\right.\\&\left.\ +3\color{blue}{\mathbf{H}_{2}\left(\frac12\right)}\ln^22-3\operatorname{Li}_3\left(\frac12\right)\ln^22+6\color{blue}{\mathbf{H}_{3}\left(\frac12\right)}\ln 2\right.\\&\ -6\operatorname{Li}_4(x)\ln 2+6\color{blue}{\mathbf{H}_{4}(x)}-6\operatorname{Li}_5(x)\bigg]\\
=&\ 12\zeta(5)-\pi^2\zeta(3)+\frac{3}8\zeta(3)\ln^22-\frac{\pi^4}{120}\ln2-\frac{1}
{4}\ln^52\\&\ -6\color{blue}{\mathbf{H}_{4}\left(\frac12\right)}+6\operatorname{Li}_4\left(\frac12\right)\ln 2+6\operatorname{Li}_5\left(\frac12\right).
\end{align}
Using the similar approach as calculating the blue integral, then
\begin{align}
\int\frac{\ln^3x\ln (1-x)}{x}\ dx&=-\int\sum_{n=1}^\infty \frac{x^{n-1}}{n}\ln^3x\ dx\\
&=-\sum_{n=1}^\infty \frac{1}{n}\int x^{n-1}\ln^3x\ dx\\
&=-\sum_{n=1}^\infty \frac{1}{n}\frac{\partial^3}{\partial n^3}\left[\int x^{n-1}\ dx\right]\\
&=-\sum_{n=1}^\infty \frac{1}{n}\frac{\partial^3}{\partial n^3}\left[\frac{x^{n}}{n}\right]\\
&=-\sum_{n=1}^\infty \frac{1}{n}\left[\frac{x^{n}\ln^3x}{n}-\frac{3x^{n}\ln^2x}{n^2}+\frac{6x^{n}\ln x}{n^3}-\frac{6x^{n}}{n^4}\right]\\
&=\sum_{n=1}^\infty \left[-\frac{x^{n}\ln^3x}{n^2}+\frac{3x^{n}\ln^2x}{n^3}-\frac{6x^{n}\ln x}{n^4}+\frac{6x^{n}}{n^5}\right]\\
&=6\operatorname{Li}_5(x)-6\operatorname{Li}_4(x)\ln x+3\operatorname{Li}_3(x)\ln^2x-\operatorname{Li}_2(x)\ln^3x.
\end{align}
Hence
$$
\int_{\large\frac{1}{2}}^1\frac{\ln^3x\ln (1-x)}{x}\ dx=\frac{\pi^2}{6}\ln^32-\frac{21}{8}\zeta(3)\ln^22-6\operatorname{Li}_4\left(\frac{1}{2}\right)\ln2-6\operatorname{Li}_5\left(\frac{1}{2}\right)+6\zeta(5).
$$
Combining altogether, we have
\begin{align}
I=&\ \frac{\pi^4}{120}\ln2-\frac{33}4\zeta(3)\ln^22+\frac{\pi^2}2\ln^32-\frac{11}{20}\ln^52+6\zeta(5)+\pi^2\zeta(3)\\
&\ +6\color{blue}{\mathbf{H}_{4}\left(\frac12\right)}-18\operatorname{Li}_4\left(\frac12\right)\ln2-24\operatorname{Li}_5\left(\frac12\right).
\end{align}
Continuing my answer in: A sum containing harmonic numbers $\displaystyle\sum_{n=1}^\infty\frac{H_n}{n^3\,2^n}$, we have
\begin{align}
\color{blue}{\mathbf{H}_{3}\left(x\right)}=&\frac12\zeta(3)\ln x-\frac18\ln^2x\ln^2(1-x)+\frac12\ln x\left[\color{blue}{\mathbf{H}_{2}\left(x\right)}-\operatorname{Li}_3(x)\right]\\&+\operatorname{Li}_4(x)-\frac{\pi^2}{12}\operatorname{Li}_2(x)-\frac12\operatorname{Li}_3(1-x)\ln x+\frac{\pi^4}{60}.\tag1
\end{align}
Dividing $(1)$ by $x$ and then integrating yields
$$\small\begin{align}
\color{blue}{\mathbf{H}_{4}\left(x\right)}=&\frac14\zeta(3)\ln^2 x-\frac18\int\frac{\ln^2x\ln^2(1-x)}x\ dx+\frac12\int\frac{\ln x}x\bigg[\color{blue}{\mathbf{H}_{2}\left(x\right)}-\operatorname{Li}_3(x)\bigg]\ dx\\&+\operatorname{Li}_5(x)-\frac{\pi^2}{12}\operatorname{Li}_3(x)-\frac12\int\frac{\operatorname{Li}_3(1-x)\ln x}x\ dx+\frac{\pi^4}{60}\ln x\\
=&\frac14\zeta(3)\ln^2 x+\frac{\pi^4}{60}\ln x+\operatorname{Li}_5(x)-\frac{\pi^2}{12}\operatorname{Li}_3(x)-\frac18\color{red}{\int\frac{\ln^2x\ln^2(1-x)}x\ dx}\\&+\frac12\left[\color{purple}{\sum_{n=1}^\infty\frac{H_{n}}{n^2}\int x^{n-1}\ln x\ dx}-\color{green}{\int\frac{\operatorname{Li}_3(x)\ln x}x\ dx}-\color{orange}{\int\frac{\operatorname{Li}_3(1-x)\ln x}x\ dx}\right].\tag2
\end{align}$$
Evaluating the red integral using the same technique as the previous one yields
\begin{align}
\color{red}{\int\frac{\ln^2x\ln^2(1-x)}x\ dx}&=\frac13\ln^3x\ln^2(1-x)-\frac23\color{blue}{\int\frac{\ln(1-x)\ln^3 x}{1-x}\ dx}.
\end{align}
Evaluating the purple integral yields
\begin{align}
\color{purple}{\sum_{n=1}^\infty\frac{H_{n}}{n^2}\int x^{n-1}\ln x\ dx}&=\sum_{n=1}^\infty\frac{H_{n}}{n^2}\frac{\partial}{\partial n}\left[\int x^{n-1}\ dx\right]\\
&=\sum_{n=1}^\infty\frac{H_{n}}{n^2}\left[\frac{x^n\ln x}{n}-\frac{x^n}{n^2}\right]\\
&=\color{blue}{\mathbf{H}_{3}(x)}\ln x-\color{blue}{\mathbf{H}_{4}(x)}.
\end{align}
Evaluating the green integral using IBP by setting $u=\ln x$ and $dv=\dfrac{\operatorname{Li}_3(x)}{x}\ dx$ yields
\begin{align}
\color{green}{\int\frac{\operatorname{Li}_3(x)\ln x}x\ dx}&=\operatorname{Li}_4(x)\ln x-\int\frac{\operatorname{Li}_4(x)}x\ dx\\
&=\operatorname{Li}_4(x)\ln x-\operatorname{Li}_5(x).
\end{align}
Evaluating the orange integral using IBP by setting $u=\operatorname{Li}_3(1-x)$ and $dv=\dfrac{\ln x}{x}\ dx$ yields
\begin{align}
\color{orange}{\int\frac{\operatorname{Li}_3(1-x)\ln x}x\ dx}&=\frac12\operatorname{Li}_3(1-x)\ln^2 x+\frac12\color{maroon}{\int\frac{\operatorname{Li}_2(1-x)\ln^2 x}{1-x}\ dx}.
\end{align}
Applying IBP again to evaluate the maroon integral by setting $u=\operatorname{Li}_2(1-x)$ and
$$
dv=\dfrac{\ln^2 x}{1-x}\ dx\quad\Rightarrow\quad
v=2\operatorname{Li}_3(x)-2\operatorname{Li}_2(x)\ln x-\ln(1-x)\ln^2x,
$$
we have
$$\small{\begin{align}
\color{maroon}{\int\frac{\operatorname{Li}_2(1-x)\ln^2 x}{1-x}\ dx}=&\left[2\operatorname{Li}_3(x)-2\operatorname{Li}_2(x)\ln x-\ln(1-x)\ln^2x\right]\operatorname{Li}_2(1-x)\\
&-2\int\frac{\operatorname{Li}_3(x)\ln x}{1-x}\ dx+2\int\frac{\operatorname{Li}_2(x)\ln x}{1-x}\ dx+\color{blue}{\int\frac{\ln(1-x)\ln^3 x}{1-x}\ dx}.
\end{align}}$$
We use the generating function for the generalized harmonic numbers evaluate the above integrals involving polylogarithm.
\begin{align}
\int\frac{\operatorname{Li}_k(x)\ln x}{1-x}\ dx&=\sum_{n=1}^\infty H_{n}^{(k)}\int x^n\ln x\ dx\\
&=\sum_{n=1}^\infty H_{n}^{(k)}\frac{\partial}{\partial n}\left[\int x^n\ dx\right]\\
&=\sum_{n=1}^\infty H_{n}^{(k)}\left[\frac{x^{n+1}\ln x}{n+1}-\frac{x^{n+1}}{(n+1)^2}\right]\\
&=\sum_{n=1}^\infty\left[\frac{H_{n+1}^{(k)}x^{n+1}\ln x}{n+1}-\frac{x^{n+1}\ln x}{(n+1)^{k+1}}-\frac{H_{n+1}^{(k)}x^{n+1}}{(n+1)^2}+\frac{x^{n+1}}{(n+1)^{k+2}}\right]\\
&=\sum_{n=1}^\infty\left[\frac{H_{n}^{(k)}x^{n}\ln x}{n}-\frac{x^{n}\ln x}{n^{k+1}}-\frac{H_{n}^{(k)}x^{n}}{n^2}+\frac{x^{n}}{n^{k+2}}\right]\\
&=\color{blue}{\mathbf{H}_{1}^{(k)}(x)}\ln x-\operatorname{Li}_{k+1}(x)\ln x-\color{blue}{\mathbf{H}_{2}^{(k)}(x)}+\operatorname{Li}_{k+2}(x).
\end{align}
Dividing generating function of $\color{blue}{\mathbf{H}^{(k)}(x)}$ by $x$ and then integrating yields
\begin{align}
\sum_{n=1}^\infty \frac{H_{n}^{(k)}x^n}{n}&=\int\frac{\operatorname{Li}_k(x)}{x(1-x)}\ dx\\
\color{blue}{\mathbf{H}_{1}^{(k)}(x)}&=\int\frac{\operatorname{Li}_k(x)}{x}\ dx+\int\frac{\operatorname{Li}_k(x)}{1-x}\ dx\\
&=\operatorname{Li}_{k+1}(x)+\int\frac{\operatorname{Li}_k(x)}{1-x}\ dx.
\end{align}
Repeating the process above yields
\begin{align}
\sum_{n=1}^\infty \frac{H_{n}^{(k)}x^n}{n^2}
&=\int\frac{\operatorname{Li}_{k+1}(x)}{x}\ dx+\int\frac{\operatorname{Li}_k(x)}{x(1-x)}\ dx\\
\color{blue}{\mathbf{H}_{2}^{(k)}(x)}&=\operatorname{Li}_{k+2}(x)+\operatorname{Li}_{k+1}(x)+\int\frac{\operatorname{Li}_k(x)}{1-x}\ dx,
\end{align}
where it is easy to show by using IBP that
\begin{align}
\int\frac{\operatorname{Li}_2(x)}{1-x}\ dx&=-\int\frac{\operatorname{Li}_2(1-x)}{x}\ dx\\
&=2\operatorname{Li}_3(x)-2\operatorname{Li}_2(x)\ln(x)-\operatorname{Li}_2(1-x)\ln x-\ln (1-x)\ln^2x
\end{align}
and
$$
\int\frac{\operatorname{Li}_3(x)}{1-x}\ dx=-\int\frac{\operatorname{Li}_3(1-x)}{x}\ dx=-\frac12\operatorname{Li}_2^2(1-x)-\operatorname{Li}_3(1-x)\ln x.
$$
Now, all unknown terms have been obtained. Putting altogether to $(2)$, we have
$$\small{\begin{align}
\color{blue}{\mathbf{H}_{4}(x)}
=&\ \frac1{10}\zeta(3)\ln^2 x+\frac{\pi^4}{150}\ln x-\frac{\pi^2}{30}\operatorname{Li}_3(x)-\frac1{60}\ln^3x\ln^2(1-x)+\frac65\operatorname{Li}_5(x)\\&-\frac15\left[\operatorname{Li}_3(x)-\operatorname{Li}_2(x)\ln x-\frac12\ln(1-x)\ln^2x\right]\operatorname{Li}_2(1-x)-\frac15\operatorname{Li}_4(x)\\&-\frac35\operatorname{Li}_4(x)\ln x+\frac15\operatorname{Li}_3(x)\ln x+\frac15\operatorname{Li}_3(x)\ln^2x-\frac1{10}\operatorname{Li}_3(1-x)\ln^2 x\\&-\frac1{15}\operatorname{Li}_2(x)\ln^3x-\frac15\color{blue}{\mathbf{H}_{2}^{(3)}(x)}+\frac15\color{blue}{\mathbf{H}_{2}^{(2)}(x)}
+\frac15\color{blue}{\mathbf{H}_{1}^{(3)}(x)}\ln x\\&-\frac15\color{blue}{\mathbf{H}_{1}^{(2)}(x)}\ln x+\frac25\color{blue}{\mathbf{H}_{3}(x)}\ln x-\frac15\color{blue}{\mathbf{H}_{2}(x)}\ln^2x+\frac1{15}\color{blue}{\mathbf{H}_{1}(x)}\ln^3x+C.\tag3
\end{align}}$$
The next step is finding the constant of integration. Setting $x=1$ to $(3)$ yields
$$\small{\begin{align}
\color{blue}{\mathbf{H}_{4}(1)}
&=-\frac{\pi^2}{30}\operatorname{Li}_3(1)+\frac65\operatorname{Li}_5(1)-\frac15\operatorname{Li}_4(1)-\frac15\color{blue}{\mathbf{H}_{2}^{(3)}(1)}+\frac15\color{blue}{\mathbf{H}_{2}^{(2)}(1)}+C\\
3\zeta(5)+\zeta(2)\zeta(3)&=-\frac{\pi^2}{30}\operatorname{Li}_3(1)+\frac{19}{30}\operatorname{Li}_5(1)+\frac{3}{5}\operatorname{Li}_3(1)+C\\
C&=\frac{\pi^4}{450}+\frac{\pi^2}{5}\zeta(3)-\frac35\zeta(3)+3\zeta(5).
\end{align}}$$
Thus
$$\small{\begin{align}
\color{blue}{\mathbf{H}_{4}(x)}
=&\ \frac1{10}\zeta(3)\ln^2 x+\frac{\pi^4}{150}\ln x-\frac{\pi^2}{30}\operatorname{Li}_3(x)-\frac1{60}\ln^3x\ln^2(1-x)+\frac65\operatorname{Li}_5(x)\\&-\frac15\left[\operatorname{Li}_3(x)-\operatorname{Li}_2(x)\ln x-\frac12\ln(1-x)\ln^2x\right]\operatorname{Li}_2(1-x)-\frac15\operatorname{Li}_4(x)\\&-\frac35\operatorname{Li}_4(x)\ln x+\frac15\operatorname{Li}_3(x)\ln x+\frac15\operatorname{Li}_3(x)\ln^2x-\frac1{10}\operatorname{Li}_3(1-x)\ln^2 x\\&-\frac1{15}\operatorname{Li}_2(x)\ln^3x-\frac15\color{blue}{\mathbf{H}_{2}^{(3)}(x)}+\frac15\color{blue}{\mathbf{H}_{2}^{(2)}(x)}
+\frac15\color{blue}{\mathbf{H}_{1}^{(3)}(x)}\ln x\\&-\frac15\color{blue}{\mathbf{H}_{1}^{(2)}(x)}\ln x+\frac25\color{blue}{\mathbf{H}_{3}(x)}\ln x-\frac15\color{blue}{\mathbf{H}_{2}(x)}\ln^2x+\frac1{15}\color{blue}{\mathbf{H}_{1}(x)}\ln^3x\\&+\frac{\pi^4}{450}+\frac{\pi^2}{5}\zeta(3)-\frac35\zeta(3)+3\zeta(5)\tag4
\end{align}}$$
and setting $x=\frac12$ to $(4)$ yields
\begin{align}
\color{blue}{\mathbf{H}_{4}\left(\frac12\right)}=&\ \frac{\ln^52}{40}-\frac{\pi^2}{36}\ln^32+\frac{\zeta(3)}{2}\ln^22-\frac{\pi^2}{12}\zeta(3)\\&+\frac{\zeta(5)}{32}-\frac{\pi^4}{720}\ln2+\operatorname{Li}_4\left(\frac12\right)\ln2+2\operatorname{Li}_5\left(\frac12\right).\tag5
\end{align}
Finally, we obtain
\begin{align}
\int_0^1\frac{\ln^3(1+x)\ln x}x\ dx=&\ \color{blue}{\frac{\pi^2}2\zeta(3)+\frac{99}{16}\zeta(5)-\frac25\ln^52+\frac{\pi^2}3\ln^32-\frac{21}4\zeta(3)\ln^22}\\&\color{blue}{-12\operatorname{Li}_4\left(\frac12\right)\ln2-12\operatorname{Li}_5\left(\frac12\right)},
\end{align}
which again matches @Cleo's answer.
References :
$[1]\ $ Harmonic number
$[2]\ $ Polylogarithm
Using the well-known identity:
$${\rm{L}}{{\rm{i}}_3}(\frac{{ - x}}{{1 - x}}) + {\rm{L}}{{\rm{i}}_3}(1 - x) + {\rm{L}}{{\rm{i}}_3}(x) = \zeta (3) + \frac{{{\pi ^2}}}{6}\ln (1 - x) - \frac{1}{2}\ln x{\ln ^2}(1 - x) + \frac{1}{6}{\ln ^3}(1 - x)$$
we obtain (the integral on RHS can be easily evaluated by differentiating beta function):
$$2\int_0^1 {\frac{{{\rm{L}}{{\rm{i}}_3}(1 - x)}}{{\sqrt {x(1 - x)} }}dx} + \int_0^1 {\frac{{{\rm{L}}{{\rm{i}}_3}(\frac{{ - x}}{{1 - x}})}}{{\sqrt {x(1 - x)} }}dx} = - 2\pi \zeta (3) + \frac{8}{3}\pi {\ln ^3}2 - \frac{2}{3}{\pi ^3}\ln 2$$
By transformation $u=x/(1-x)$, we have
$$\int_0^1 {\frac{{{\rm{L}}{{\rm{i}}_3}(\frac{{ - x}}{{1 - x}})}}{{\sqrt {x(1 - x)} }}dx} = \int_0^\infty {\frac{{{\rm{L}}{{\rm{i}}_3}( - u)}}{{(1 + u)\sqrt u }}du}$$
I claim this integral is $-6\pi \zeta(3)$.
To establish this value, it suffices to show, with $\zeta(\cdot,\cdot)$ Hurwitz zeta function,
$$\int_0^\infty {\frac{{{\rm{L}}{{\rm{i}}_3}( - x)}}{{1 + x}}{x^{s - 1}}dx} = \frac{\pi }{{\sin (\pi s)}}\left[ {\zeta (3) - \zeta (3,1 - s)} \right] \qquad 0<s<1$$
by Mellin inversion theorem, this in turn is equivalent to, (which applies as the function tends to $0$ uniformly in the vertical strip $0<\Re(s)<1$ thanks to the $\csc(s\pi)$ factor) for an instance of $c$ with $0<c<1$:
$$\tag{1} \frac{{{\rm{L}}{{\rm{i}}_3}( - x)}}{{1 + x}} = \frac{1}{{2\pi i}}\int_{c - i\infty }^{c + i\infty } {\frac{{\pi {x^{ - s}}}}{{\sin (\pi s)}}\left[ {\zeta (3) - \zeta (3,1 - s)} \right]ds} \qquad x>0$$
Note that both sides of $(1)$ is an analytic function for $\Re(x) > 0$, hence it suffices to consider the case when $0<x<1$. When this is the case, we can draw a vertical semicircle on the left half-plane, with vertices $c \pm i\infty$, then the integral on the semicircle tends to $0$, calculating residues at $-1,-2,\cdots$ gives
$$\begin{aligned}\frac{1}{{2\pi i}}\int_{c - i\infty }^{c + i\infty } {\frac{{\pi {x^{ - s}}}}{{\sin (\pi s)}}\left[ {\zeta (3) - \zeta (3,1 - s)} \right]ds} &= \sum\limits_{n = 1}^\infty {{{( - x)}^n}\left[ {\zeta (3) - \zeta (3,1 + n)} \right]} \\ &=\sum\limits_{n = 1}^\infty {{{( - x)}^n}\sum\limits_{k = 1}^n {\frac{1}{{{k^3}}}} } = \frac{{{\rm{L}}{{\rm{i}}_3}( - x)}}{{1 + x}}
\end{aligned}$$
where we exchanged two order of summations, completing the proof.
Best Answer
\begin{align} &\int_0^1\ln^n\left(\frac{1-x}{1+x}\right)\overset{x\to \frac{1-x}{1+x} }{dx}\\ =&\ 2\int_0^1\frac{\ln^n x}{(1+x)^2}dx= 2\int_0^1\ln^n x \ d\left(\frac{x}{1+x}\right)\\ \overset{ibp}=& -2n\int_0^1\frac{\ln^{n-1} x}{1+x}dx =-2n\left(\int_0^1 \frac{\ln^{n-1} x}{1-x} {d}x-\int_0^1 \frac{2x\ln^{n-1} x}{1-x^2} \overset{x^2\to x}{dx}\right)\\ =&-2n \left(1- \frac1{2^{n-1}}\right) \int_0^1 \frac{\ln^{n-1} x}{1-x} {d}x = 2n!(-1)^{n}\left(1-\frac1{2^{n-1}}\right) \zeta(n) \end{align}