Closed-forms for the integral $\int_0^1\frac{\operatorname{Li}_n(x)}{1+x}dx$? (This is related to this question.)
Define the integral,
$$I_n = \int_0^1\frac{\operatorname{Li}_n(x)}{1+x}dx$$
with polylogarithm $\operatorname{Li}_n(x)$. Given the Nielsen generalized polylogarithm $S_{n,p}(z)$,
$$S_{n,p}(z) = \frac{(-1)^{n+p-1}}{(n-1)!\,p!}\int_0^1\frac{(\ln t)^{n-1}\big(\ln(1-z\,t)\big)^p}{t}dt$$
Then it seems,
$$I_1 = -S_{1,1}(-1)-\tfrac12\ln(2)\ln(2)$$
$$I_2 = -5S_{1,2}(-1)+\ln(2)\,\zeta(2)\quad$$
$$\quad\qquad I_3 = -2S_{1,3}(-1)+\ln(2)\,\zeta(3)-\tfrac12\zeta(4)$$
where $S_{1,1}(-1) = -\tfrac12\zeta(2)$, and $S_{1,2}(-1) = \tfrac18\zeta(3)$ and $S_{1,3}(-1)$ has a more complicated closed-form given in the linked post.

Q: What is $I_4$ and $I_5$? In general, can $I_n$ be expressed by the Nielsen generalized polylogarithm?

P.S. Note that $\operatorname{Li}_n(z), \ln(z), \zeta(z)$ are just special cases of this function.
 A: Based on Three-sided coin's comments, it seems
$$I_4 = \int_0^1 \frac{\operatorname{Li}_4(x)}{1+x}dx =\ln(2)\zeta(4)+\tfrac34\zeta(2)\zeta(3)-\tfrac{59}{32}\zeta(5) \approx 0.321352\dots$$
(Note: Confirmed as correct by Brevan's answer.) Connecting it to Three-sided coin's other integrals, then
$$I_4 = \ln(2)\zeta(4)-\tfrac12\zeta(2)\zeta(3)-h_1$$
$$I_4 = \ln(2)\zeta(4)+\tfrac14\zeta(2)\zeta(3)-h_2$$
where,
$$h_1=\int_0^1\frac{\operatorname{Li}_2(x)\,\color{blue}{\operatorname{Li}_2(-x)}}{x}dx = -\tfrac54\zeta(2)\zeta(3)+\tfrac{59}{32}\zeta(5)$$
$$h_2=\int_0^1\frac{\color{blue}{\operatorname{Li}_3(-x)}\ln(1-x)}{x}dx = -\tfrac12\zeta(2)\zeta(3)+\tfrac{59}{32}\zeta(5)$$
which implies,
$$h_1-h_2 =\int_0^1\frac{\operatorname{Li}_2(x)\,\operatorname{Li}_2(-x)}{x}dx- \int_0^1\frac{\operatorname{Li}_3(-x)\ln(1-x)}{x}dx =-\tfrac34\zeta(2)\zeta(3)$$

Compare to the similar integrals here that he mentioned,
$$h_3= \int_0^1\frac{\operatorname{Li}_2(x)\,\color{blue}{\operatorname{Li}_2(x)}}{x}dx = 2\zeta(2)\zeta(3)-3\zeta(5)$$
$$h_4 = \int_0^1\frac{\color{blue}{\operatorname{Li}_3(x)}\ln(1-x)}{x}dx =\zeta(2)\zeta(3)-3\zeta(5$$
which has the proven relation,
$$h_3-h_4 = \int_0^1\frac{\operatorname{Li}_2(x)\,\operatorname{Li}_2(x)}{x}dx - \int_0^1\frac{\operatorname{Li}_3(x)\ln(1-x)}{x}dx  = \zeta(2)\zeta(3)$$


Update: Per Brevan's answer:

$$I_n = \int_0^1\frac{\operatorname{Li}_n(x)}{1+x}dx=(-1)^nA(n)+\sum_{k=1}^{n-1}(-1)^{k+1}\eta(k)\zeta(n+1-k)$$
with Dirichlet eta function $\eta(k)$ and
$$A(n) = \sum_{k=1}^\infty \frac{H_k}{k^n}(-1)^k$$
is the $n$th "Alternating Euler Sum". However, since,
$$A(n,z) = \sum_{k=1}^\infty \frac{H_k}{k^n}z^k =S_{n-1,2}(z)+\operatorname{Li}_{n+1}(z)$$
for $-1\leq z\leq 1$, then $I_n$ can be expressed by Nielsen polylogs $S_{n,p}(z)$ as I suspected.
A: Per @TitoPiezas' request, I will expand on my comment above and essentially close this question as far as the literature we have (unless there are new or unknown developments I am unaware of).
As @Zacky said, a key step in converting this problem to a more conventional one is repeated integration by parts. Using the techniques outlined in the comments above, we get
$$I_n=\sum_{k=1}^{n-1}(-1)^{k+1}\eta(k)\zeta(n+1-k)+\sum_{k=0}^\infty \tfrac{(-1)^{k+n}}{(k+1)^{n-1}}\int_0^1x^k\text{Li}_1(x) dx$$
We now shift the right summation index and use $\text{Li}_1(x) = -\log(1-x)$ to get 
$$I_n=\sum_{k=1}^{n-1}(-1)^{k+1}\eta(k)\zeta(n+1-k)+\sum_{k=1}^\infty \tfrac{(-1)^{k+n}}{k^{n-1}}\int_0^1x^{k-1}\log(1-x) dx$$
and recall that$$\int_0^1x^{k-1}\log(1-x) dx= \frac{H_k}{k}$$
So we finally conclude
$$I_n=\sum_{k=1}^{n-1}(-1)^{k+1}\eta(k)\zeta(n+1-k)+(-1)^nA(n)$$
where $A(n) = \sum_{k=1}^\infty \tfrac{(-1)^k H_k}{k^n}$ is the $n$th "Alternating Euler Sum". 
Alternating Euler Sums are well studied; examples just from this website can be found at this basic survey of integral forms here and the most amazing collection of answers ever here.
That latter link, when converted over to the more simple notation outlined here, gives a closed form for $A(2n)$ and shows why $A(2n+1)$ is particularly tricky; using the results on that page we should be able to easily get $A(1)$ and $A(3)$ but once we move higher we get linear dependence chains in the recurrence formulas that seemingly prevent us from expressing our answers in terms of Polylogarithms $\text{Li}_s(z)$ (which includes the Zeta and Dirichlet Eta functions as special values at $z=\pm 1$ respectively) and elementary functions in general (though specific cases, in particular for low $n$, may still work).

Note: I cannot stress enough how amazing that second link is. It essentially settles this question on its own through the amazing answers! Please take the time to go upvote those answers if possible.
A: Here is a generalization thats works for even $n$:
\begin{gather*}
\int_0^1\frac{\operatorname{Li}_{2a}(x)}{1+x}\mathrm{d}x=\int_0^1\frac{\operatorname{Li}_{2a}(x)}{x}\mathrm{d}x-\int_0^1\frac{\operatorname{Li}_{2a}(x)}{x(1+x)}\mathrm{d}x\\
=\operatorname{Li}_{2a+1}(x)|_0^1-\int_0^1\frac{1}{1+x}\left(\frac{-1}{(2a-1)!}\int_0^1\frac{\ln^{2a-1}(y)}{1-xy}\mathrm{d}y\right)\mathrm{d}x\\
=\operatorname{Li}_{2a+1}(1)+\frac{1}{(2a-1)!}\int_0^1\ln^{2a-1}(y)\left(\int_0^1\frac{\mathrm{d}x}{(1+x)(1-yx)}\right)\mathrm{d}y\\
=\zeta(2a+1)+\frac{1}{(2a-1)!}\int_0^1\ln^{2a-1}(y)\left(\frac{\ln(2)-\ln(1-y)}{1+y}\right)\mathrm{d}y\\
=\zeta(2a+1)+\frac{\ln(2)}{(2a-1)!}\int_0^1\frac{\ln^{2a-1}(y)}{1+y}\mathrm{d}y\\
-\frac{1}{(2a-1)!}\int_0^1\frac{\ln^{2a-1}(y)\ln(1-y)}{1+y}\mathrm{d}y.
\end{gather*}
Substitute
$$\int_0^1\frac{\ln^a(x)}{1+x}\mathrm{d}x=(-1)^aa!\,(1-2^{-a})\zeta(a+1)$$
and (this integral is evaluated by Cornel here)
\begin{gather*}
\int_0^1\frac{\ln^{2q-1}(x)\ln(1-x)}{1+x}\mathrm{d}x=(2^{1-2q}-2)(2q-1)!\ln(2)\zeta(2q)\\
+[q-2^{-1-2q}(1+2q-2^{1+2q})](2q-1)!\,\zeta(2q+1)\\
-(2q-1)!\sum_{k=1}^{q-1}(1-2^{2k-2q})\zeta(2k)\zeta(2q-2k+1),
\end{gather*}
we get
\begin{gather*}
\int_0^1\frac{\operatorname{Li}_{2a}(x)}{1+x}\mathrm{d}x=[1-a+2^{-1-2a}(2a+1-2^{1+2a})]\zeta(2a+1)\nonumber\\
+\ln(2)\zeta(2a)+\sum_{k=1}^{a-1}(1-2^{2k-2a})\zeta(2k)\zeta(2a-2k+1).
\end{gather*}

Examples
\begin{gather}
\int_0^1\frac{\operatorname{Li}_{2}(x)}{1+x}\mathrm{d}x=\ln(2)\zeta(2)-\frac58\zeta(3);\\
\int_0^1\frac{\operatorname{Li}_{4}(x)}{1+x}\mathrm{d}x=\ln(2)\zeta(4)+\frac34\zeta(2)\zeta(3)-\frac{59}{32}\zeta(5)\label{int_0^1 Li_4(x)/(1+x)};\\
\int_0^1\frac{\operatorname{Li}_{6}(x)}{1+x}\mathrm{d}x=\ln(2)\zeta(6)+\frac34\zeta(3)\zeta(4)+\frac{15}{16}\zeta(2)\zeta(5)-\frac{377}{128}\zeta(7).
\end{gather}
