To establish notation the polylogarithm Li$_n(x)$ has the power series expansion
$$ \text{Li}_n(x)= \sum_{k=1}^\infty \frac{x^k}{k^n} $$
and the Riemann zeta can be considered the special value $\zeta(n) =\text{Li}_n(1).$
Define
$$F_1(r)=\text{Li}_1(r)=-\log(1-r)$$
$$F_2(r)=\int_0^r dt \big(\frac{1}{t} + \frac{2}{1-t} \big) F_1(t) = (\text{Li}_1(r))^2 + \text{Li}_2(r) $$
$$F_3(r)=\int_0^r dt \big(\frac{1}{t} + \frac{2}{1-t} \big) F_2(t) = $$
$$ =\frac{2}{3}(\text{Li}_1(r))^3 + \text{Li}_1(r)\big(\text{Li}_2(r)+ \text{Li}_2(1-r)+\zeta(2)\big) + \text{Li}_3(r)+ 2\text{Li}_3(1-r) – 2\zeta(3) $$
where the closed forms can be found with Mathematica, although manual simplification was carried out for the last expression. In general, define
$$F_n(r)=\int_0^r dt \big(\frac{1}{t} + \frac{2}{1-t} \big) F_{n-1}(t)$$
Question 1: Can explicit forms be found for $F_4$ or for larger $n$?
Question 2: Can it be shown that
$$ \int_0^1 F_n(-r) \frac{dr}{r} = -\frac{\zeta(n+1)}{2} \quad \text{ and } \quad \int_0^1 F_n(+r) \frac{dr}{r}
= \big(2^n-1)\zeta{(n+1)} $$
(In addition to the closed forms above, for $n=4$ I get numerical agreement. )
Best Answer
I have found $F_4$ (Question 1) and performed a numerical integration to confirm Question 2 to the next degree. Let $L_1:=\text{Li}_1(r).$ Then
$$F_4(r)=\frac{(L_1)^4}{4} + \frac{(L_1)^3}{3} \log{r} + (L_1)^2 \zeta(2) - 2L_1\zeta(3)-2\text{Li}_4(1-r) - \text{Li}_4(r) - 2\text{Li}_4(\frac{-r}{1-r}) + 2\zeta(4) $$
The technique was not profound, but used Mathematica to do the integration, once $F_3$ was put into a different form using dilogarithm identities (see below). However, the form by Math. had over twenty terms, and many of those involved powers of $\log{(r/(1-r))} = \log{r} - \log{(1-r)} .$ Using that logarithm property and extensive simplification with the repeated use of $$ \text{Li}_2(x) + \text{Li}_2(1-x) = \zeta(2) - \log{x}\log{(1-x)} $$ $$ \text{Li}_2(x) + \text{Li}_2(-x/(1-x)) = - \frac{1}{2}\log^2{(1-x)} $$ $$ \text{Li}_3(x) + \text{Li}_3(1-x) + \text{Li}_3(-x/(1-x)) = \zeta(2)\log{(1-x)} - \frac{1}{2}\log{x}\log^2{(1-x)}+ \frac{1}{6} \log^3{(1-x)} $$
resulted in the answer of the form given. It is interesting that the final form is somewhat simple, involving no $\text{Li}_2(x)$ or $\text{Li}_3(x)$ terms where $x=r, 1-r,$ or $-r/(1-r),$ although $x=1$ does appear. Note that
$$F_3(r)=\frac{2(L_1)^3}{3} + (L_1)^2 \log{r} + 2L_1 \zeta(2) +\text{Li}_3(r) + 2(\text{Li}_3(1-r) -\zeta(3)) $$ (the form integrated) has similar characteristics: all the powers of $\log{r}$ except for that of the first degree are cancelled, and there are no terms involving no $\text{Li}_2(x)$ where $x=r, 1-r,$ or $-r/(1-r).$
I believe more can be said about this problem, and I'll award points for further progress, even though I'll accept my own answer to avoid completely losing bounty points.