$\int_0^1\frac{\ln\left(1-t\right)\ln^3 t}{2-t}dt$ From results and techniques of Integrating $\int_0^1 \frac{\ln(1+x)\ln^3 x}{1+x}\,dx$ with restricted techniques
i think i can prove that:
$\displaystyle \int_0^1\frac{\ln\left(1-t\right)\ln^3 t}{2-t}dt=-\frac{3}{16}\zeta(5)-\frac{21}{4}\zeta(2)\zeta(3)+\frac{9}{4}\zeta(4)\ln 2-\frac{3}{20}\ln^5 2+18 \text{Li}_5\left(\frac{1}{2}\right)$
I'm wondering if this result is obtainable via beta functions and/or (generalised) harmonic series.
 A: $$I=\int_0^1\frac{\ln(1-x)\ln^3x}{2-x}dx=\sum_{n=1}^\infty\frac{1}{2^n}\int_0^1 x^{n-1}\ln(1-x)\ln^3x\ dx$$
$$=\sum_{n=1}^\infty\frac{1}{2^n}\frac{\partial^3}{\partial n^3}\int_0^1 x^{n-1}\ln(1-x)\ dx$$
$$=\sum_{n=1}^\infty\frac{1}{2^n}\frac{\partial^3}{\partial n^3}\left(-\frac{H_n}{n}\right)$$
$$=6\sum_{n=1}^\infty\frac{1}{2^n}\left(\frac{H_n}{n^4}+\frac{H_n^{(2)}}{n^3}+\frac{H_n^{(3)}}{n^2}+\frac{H_n^{(4)}}{n}-\frac{\zeta(2)}{n^3}-\frac{\zeta(3)}{n^2}-\frac{\zeta(4)}{n}\right)$$
$$\small{=6\left(\sum_{n=1}^\infty\frac{H_n}{n^42^n}+\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^32^n}+\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^22^n}+\sum_{n=1}^\infty\frac{H_n^{(4)}}{n2^n}-\zeta(2)\text{Li}_3\left(\frac12\right)-\zeta(3)\text{Li}_2\left(\frac12\right)-\ln(2)\zeta(4)\right)}$$
The last three series dont need to be calculated individually:
By Cauchy product we have
$$-\ln(1-x)\text{Li}_4(x)=2 \sum_{n=1}^\infty\frac{H_n}{n^4}x^n+\sum_{n=1}^\infty \frac{H_n^{(2)}}{n^3}x^n+\sum_{n=1}^\infty \frac{H_n^{(3)}}{n^2}x^n+ \sum_{n=1}^\infty\frac{H_n^{(4)}}{n}x^n-5\text{Li}_5(x)$$
Set $x=1/2$ we have
$$\sum_{n=1}^\infty \frac{H_n^{(2)}}{n^32^n}+\sum_{n=1}^\infty \frac{H_n^{(3)}}{n^22^n}+ \sum_{n=1}^\infty\frac{H_n^{(4)}}{n2^n}=5\text{Li}_5\left(\frac12\right)+\ln(2)\text{Li}_4\left(\frac12\right)-2\sum_{n=1}^\infty\frac{H_n}{n^42^n}$$
Plugging this back in yields
$$\small{I=6\left(-\sum_{n=1}^\infty\frac{H_n}{n^42^n}+5\text{Li}_5\left(\frac12\right)+\ln(2)\text{Li}_4\left(\frac12\right)-\zeta(2)\text{Li}_3\left(\frac12\right)-\zeta(3)\text{Li}_2\left(\frac12\right)-\ln(2)\zeta(4)\right)}$$
In this link we found
\begin{align}
\displaystyle\sum_{n=1}^{\infty}\frac{H_n}{ n^42^n}&=2\operatorname{Li_5}\left( \frac12\right)+\ln2\operatorname{Li_4}\left( \frac12\right)-\frac16\ln^32\zeta(2)
+\frac12\ln^22\zeta(3)\\
&\quad-\frac18\ln2\zeta(4)- 
\frac12\zeta(2)\zeta(3)+\frac1{32}\zeta(5)+\frac1{40}\ln^52
\end{align}
Substituting this result along with using $\text{Li}_2(1/2)=\frac12\zeta(2)-\frac12\ln^22$ and $\text{Li}_3(1/2)=\frac78\zeta(3)-\frac12\ln2\zeta(2)+\frac16\ln^32$ the closed form follows.
