How to find closed form for $\int_{0}^{1}\ln^k{x}\ln(1+x+x^2+\cdots+x^n){\mathrm dx\over x}=F(k)\zeta(k+2)?$ Proposed:

$$\int_{0}^{1}\ln{x}\ln(1+x+x^2+\cdots+x^n){\mathrm dx\over x}=-{n(n+2)\over (n+1)^2}\zeta(3)\tag1$$

and 

$$\int_{0}^{1}\ln^k{x}\ln(1+x+x^2+\cdots+x^n){\mathrm dx\over x}=F(k)\zeta(k+2)\tag2$$

My try:
$$\int_{0}^{1}\ln^k{x}\ln\left({x^{n+1}-1\over x-1}\right){\mathrm dx\over x}=\tag3$$
$$\int_{0}^{1}\ln^k{x}\ln(x^{n+1}-1){\mathrm dx\over x}-\int_{0}^{1}\ln^k{x}\ln(x-1){\mathrm dx\over x}\tag4$$
How do we evaluate the closed form for $(2)?$
 A: Hint. The evaluation of the following integral is sufficient to obtain an answer to your question,
$$
I_{n,k}:=\int_{0}^{1}\ln^k{x}\ln(1-x^{n+1}){\mathrm dx\over x},\qquad k,n=1,2,\cdots.
$$ One may observe that, by the change of variable
$$
u=x^{n+1},\quad \ln u =(n+1) \cdot \ln x,\quad \frac{du}u=(n+1)\cdot \frac{dx}x,
$$
one has
$$
\begin{align}
I_{n,k}=\int_{0}^{1}\ln^k{x}\ln(1-x^{n+1}){\mathrm dx\over x}&=\frac1{(n+1)^{k+1}}\int_{0}^{1}\ln^k{u}\ln(1-u){\mathrm du\over u}
\\\\&=-\frac1{(n+1)^{k+1}}\sum_{\nu\ge1}{1\over \nu}\int_{0}^{1}u^\nu\ln^k{u}\:du
\\\\&=-\frac1{(n+1)^{k+1}}\sum_{\nu\ge1}{1\over \nu}\cdot \frac{(-1)^kk!}{(\nu+1)^{k+1}}
\\\\&=-\frac{(-1)^kk!}{(n+1)^{k+1}}\sum_{\nu\ge1} \frac{1}{\nu(\nu+1)^{k+1}}
\end{align}
$$ then the latter series can be obtained using the recurrence relation
$$
\sum_{\nu\ge1} \frac{1}{\nu(\nu+1)^{k+1}}=\sum_{\nu\ge1} \frac{1+\nu-\nu}{\nu(\nu+1)^{k+1}}=\sum_{\nu\ge1} \frac{1}{\nu(\nu+1)^{k}}-\zeta(k+1)+1,\quad k\ge1,
$$ giving
$$
\sum_{\nu\ge1} \frac{1}{\nu(\nu+1)^{k+1}}=k+1-\sum_{j=1}^k\zeta(j+1),\quad k\ge1.
$$ Finally,

$$
\int_{0}^{1}\ln^k{x}\ln(1+x+x^2+\cdots+x^n){\mathrm dx\over x}=(-1)^kk!\left(1-\frac{1}{(n+1)^{k+1}} \right)\left(k+1-\sum_{j=2}^{k+1}\zeta(j)\right).
$$

