Motivation
Consider $$I_n=\int_0^1\frac{\ln(1-\omega x)}{\bar\omega-x}\ln^nxdx=\int_0^1\frac{\omega\ln(1-\omega x)}{1-\omega x}\ln^nxdx$$ where $\omega=e^{i\pi/3}$. It is known that $I_0=\frac1{18}\pi^2$ which can be deduced by integrating directly, and $$I_1=\frac23\zeta(3)-\frac\pi3\operatorname{Cl}_2\left(\frac\pi3\right)+i\cdot\frac1{324}\pi^3,$$ where $\text{Cl}$ is the Clausen Cl function.
I evaluated $I_1$ by using known antiderivative of $\frac{\ln(x-a)\ln(x-b)}{x}$.
One thing I noticed why it is special for $\omega$ is that if we replace it with other complex numbers different from $\pm1$ and $0$, the result of $I_2$ won't be very beautiful, such as $$\int_0^1\frac{\ln(1-zx)}{1-zx}\ln^2xdx,\text{ where $z=i$}$$ The result of the latter involves polylogarithm values that can not be simplified. It equals $$\tiny4 i\Re\operatorname{Li}_4\left(\frac{1}{2}+\frac{i}{2}\right)+\frac{35 \pi \zeta (3)}{64}+\frac{35}{32} i \zeta (3) \log (2)-\frac{47 i \pi ^4}{1536}+\frac{1}{96} i \ln^4(2)-\frac{5}{192} i \pi ^2 \ln^2(2)+2\beta(4)$$
Also, a CAS knows how to handle $I_2$. It gives the result
$$I_2=\frac{23}{9720}\pi^4+i\left(\frac49\pi\zeta(3)-2\operatorname{Cl}_4\left(\frac\pi3\right)\right)$$
(CAS knows the antiderivative)
So the following question comes out:
For a general $n\in\mathbb N$, does a closed form of $I_n$ exist?
Higher to $n\ge3$, the polylog-styled antiderivative no longer exists. The method becomes invalid. But the numerically-verified closed form for the real part of $I_3$ still exist.
$$\Re I_3=\frac{43}6\zeta(5)-\frac16\pi^2\zeta(3)-2\pi\operatorname{Cl}_4\left(\frac\pi3\right)$$
A failed attempt is trying to convert $I_n$ to series form: $$\sum_{k=1}^\infty H_k\omega^{k+1}\int_0^1x^k\ln^n\frac1xdx=n!\sum_{k=1}^\infty \frac{H_k\omega^{k+1}}{k^{n+1}}$$and separating it into 6 series. No luck for continuing this method so far.
Edit: I'm not looking for a proof of the result above. I'm looking for a closed form of $I_n$.
$$I=-w \int_{0}^{1} \frac{\ln(1-wx)}{1-wx} \ln^n x dx= \int_{0}^{1}\sum_{k=1}^{\infty} w^{k+1} H_k~ x^k \ln^n x ~dx= \sum_{k=1}^{\infty} w^{k+1} H_k \int_{0}^{\infty} t^n e^{-(k+1)t} dt$$ $$ \implies I= n! \sum_{k=1}^{\infty} \frac{w^{k+1} H_k}{(k+1)^{n+1}}.$$