$int_0^1frac{lnleft(1-tright)ln^3 t}{2-t}dt$

$$I=int_0^1frac{ln(1-x)ln^3x}{2-x}dx=sum_{n=1}^inftyfrac{1}{2^n}int_0^1 x^{n-1}ln(1-x)ln^3x dx$$

$$=sum_{n=1}^inftyfrac{1}{2^n}frac{partial^3}{partial n^3}int_0^1 x^{n-1}ln(1-x) dx$$

$$=sum_{n=1}^inftyfrac{1}{2^n}frac{partial^3}{partial n^3}left(-frac{H_n}{n}right)$$

$$=6sum_{n=1}^inftyfrac{1}{2^n}left(frac{H_n}{n^4}+frac{H_n^{(2)}}{n^3}+frac{H_n^{(3)}}{n^2}+frac{H_n^{(4)}}{n}-frac{zeta(2)}{n^3}-frac{zeta(3)}{n^2}-frac{zeta(4)}{n}right)$$

$$small{=6left(sum_{n=1}^inftyfrac{H_n}{n^42^n}+sum_{n=1}^inftyfrac{H_n^{(2)}}{n^32^n}+sum_{n=1}^inftyfrac{H_n^{(3)}}{n^22^n}+sum_{n=1}^inftyfrac{H_n^{(4)}}{n2^n}-zeta(2)text{Li}_3left(frac12right)-zeta(3)text{Li}_2left(frac12right)-ln(2)zeta(4)right)}$$

The last three series dont need to be calculated individually:

By Cauchy product we have

$$-ln(1-x)text{Li}_4(x)=2 sum_{n=1}^inftyfrac{H_n}{n^4}x^n+sum_{n=1}^infty frac{H_n^{(2)}}{n^3}x^n+sum_{n=1}^infty frac{H_n^{(3)}}{n^2}x^n+ sum_{n=1}^inftyfrac{H_n^{(4)}}{n}x^n-5text{Li}_5(x)$$

Set $x=1/2$ we have

$$sum_{n=1}^infty frac{H_n^{(2)}}{n^32^n}+sum_{n=1}^infty frac{H_n^{(3)}}{n^22^n}+ sum_{n=1}^inftyfrac{H_n^{(4)}}{n2^n}=5text{Li}_5left(frac12right)+ln(2)text{Li}_4left(frac12right)-2sum_{n=1}^inftyfrac{H_n}{n^42^n}$$

Plugging this back in yields

$$small{I=6left(-sum_{n=1}^inftyfrac{H_n}{n^42^n}+5text{Li}_5left(frac12right)+ln(2)text{Li}_4left(frac12right)-zeta(2)text{Li}_3left(frac12right)-zeta(3)text{Li}_2left(frac12right)-ln(2)zeta(4)right)}$$

In this link we found

begin{align} displaystylesum_{n=1}^{infty}frac{H_n}{ n^42^n}&=2operatorname{Li_5}left( frac12right)+ln2operatorname{Li_4}left( frac12right)-frac16ln^32zeta(2) +frac12ln^22zeta(3)\ &quad-frac18ln2zeta(4)- frac12zeta(2)zeta(3)+frac1{32}zeta(5)+frac1{40}ln^52 end{align}

Substituting this result along with using $text{Li}_2(1/2)=frac12zeta(2)-frac12ln^22$ and $text{Li}_3(1/2)=frac78zeta(3)-frac12ln2zeta(2)+frac16ln^32$ the closed form follows.