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I've been exploring interesting integrals on this site and fell into a rabbit hole on one particular user with a special history of answering integrals (giving only answers). I enjoyed reading the interesting solutions of other users who tried to match their solution with the given answer.

That said, I found one question requesting the answers to 2 integrals which is answered but doesn't have a solution from any other user. I couldn't find any but here are some related questions. I thought there might be sufficient interest from others who might want to provide an full solution to this question. If it's possible, perhaps link this solution to the original question as well?

That's my motivation for why I want the question solved. Unfortunately, I don't have near enough expertise to solve it. The original question has asked that only one integral be posted at a time so here's the first one. The second is now here

$$\int_0^1\frac{\ln(1-x)}{x}\text{Li}_3\left(\frac{1 + x}{2} \right)\,dx$$

Here's also the attached answer from the user : $$\begin{align*}{\large\int}_0^1\frac{\ln(1-x)\,\operatorname{Li}_3\left(\frac{1+x}2\right)}xdx&=\frac{29\,\zeta(5)}{16}-\frac{19\pi^2}{96}\zeta(3)+\frac{5\,\zeta(3)}{16}\ln^22+\frac{\ln^52}{40}\\&-\frac{5\pi^2}{72}\ln^32+\frac{11\pi^4}{1440}\ln2-3\operatorname{Li}_5\left(\tfrac12\right).\\\end{align*}$$

user220704
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    One can exploit the point $ii)$ of Theorem $2$ here https://www.researchgate.net/publication/372830396. Of course, there are more ways to go. It's a somewhat long and boring process to have it all done. – user97357329 Feb 06 '24 at 18:25
  • Integrating by parts could work; there's a cancellation, and you can break it up into multiple integrals, allowing that cancellation where $u$ is the integrand and $dv=dx$. – Marek Feb 06 '24 at 19:43

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