<s>[INST] Can we evaluate $\int_0^1 \frac{\ln (1-x) \ln ^n x}{x} d x$ without expanding $\ln(1-x)$? For $|x|<1,$ we have
$$
\begin{aligned}
& \frac{1}{1-x}=\sum_{k=0}^{\infty} x^k \quad \Rightarrow \quad \ln (1-x)=-\sum_{k=0}^{\infty} \frac{x^{k+1}}{k+1}
\end{aligned}
$$

$$
\begin{aligned}
\int_0^1 \frac{\ln (1-x)}{x} d x & =-\sum_{k=0}^{\infty} \frac{1}{k+1} \int_0^1 x^k dx \\
& =-\sum_{k=0}^{\infty} \frac{1}{(k+1)^2} \\
& =- \zeta(2) \\
& =-\frac{\pi^2}{6}
\end{aligned}
$$
$$
\begin{aligned}
\int_0^1 \frac{\ln (1-x) \ln x}{x} d x & =-\sum_{k=0}^{\infty} \frac{1}{k+1} \int_0^1 x^k \ln xdx \\
& =\sum_{k=0}^{\infty} \frac{1}{k+1}\cdot\frac{1}{(k+1)^2} \\
& =\zeta(3) \\
\end{aligned}
$$

$$
\begin{aligned}
\int_0^1 \frac{\ln (1-x) \ln ^2 x}{x} d x & =-\sum_{k=0}^{\infty} \frac{1}{k+1} \int_0^1 x^k \ln ^2 xdx \\
& =-\sum_{k=0}^{\infty} \frac{1}{k+1} \cdot \frac{2}{(k+1)^3} \\
& =-2 \zeta(4) \\
& =-\frac{\pi^4}{45}
\end{aligned}
$$

In a similar way, I dare guess that
$$\int_0^1 \frac{\ln (1-x) \ln ^n x}{x} d x  =(-1)^{n+1}\Gamma(n)\zeta(n+2),$$
where $n$ is a non-negative real number.
Proof:
$$
\begin{aligned}
\int_0^1 \frac{\ln (1-x) \ln ^n x}{x} d x & =-\sum_{k=0}^{\infty} \frac{1}{k+1} \int_0^1 x^k \ln ^n xdx \\
\end{aligned}
$$
Letting $y=-(k+1)\ln x $ transforms the last integral into a Gamma function as
$$
\begin{aligned}
\int_0^1 x^k \ln ^n x d x & =\int_{\infty}^0 e^{-\frac{k}{k+1}}\left(-\frac{y}{k+1}\right)^n\left(-\frac{1}{k+1} e^{-\frac{y}{k+1}} d y\right) \\
& =\frac{(-1)^n}{(k+1)^{n+1}} \int_0^{\infty} e^{-y} y^n d y \\
& =\frac{(-1)^n \Gamma(n+1)}{(k+1)^{n+1}}
\end{aligned}
$$
Now we can conclude that
$$
\begin{aligned}
\int_0^1 \frac{\ln (1-x) \ln ^n x}{x} d x & =(-1)^{n+1} \Gamma(n+1) \sum_{k=0}^{\infty} \frac{1}{(k+1)^{n+2}} \\
& =(-1)^{n+1} \Gamma(n+1)\zeta(n+2)
\end{aligned}
$$
Can we evaluate $\int_0^1 \frac{\ln (1-x) \ln ^n x}{x} d x$ without expanding $\ln (1-x)$?
Your comments and alternative methods are highly appreciated?
 [/INST] For natural number $n$, I can use differentiation of the integral
$$
\begin{aligned}
I(a) & =\int_0^1 x^a \ln (1-x) d x \\
& =-\int_0^1 x^a \sum_{k=0}^{\infty} \frac{1}{k+1}  x^{k+1} d x \\
& =-\sum_{k=0}^{\infty} \frac{1}{k+1} \int_0^1 x^{a+k+1} d x \\
& =-\sum_{k=0}^{\infty}\left(\frac{1}{k+1} \cdot \frac{1}{a+k+2}\right)
\end{aligned}
$$
Differentiating $I(a)$ w.r.t. $a$ by $n$ times at $a=-1$ yields our integral.
$$
\begin{aligned}
I & =-\left.\sum_{k=0}^{\infty} \frac{1}{k+1} \cdot \frac{(-1)^n n !}{(a+k+2)^{n+1}}\right|_{a=-1} \\
& =(-1)^{n+1} n ! \sum_{k=0}^{\infty} \frac{1}{k+1} \cdot \frac{1}{(k+1)^{n+1}} \\
& =(-1)^{n+1} n ! \zeta(n+2)
\end{aligned}
$$
</s>