<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] Without expanding using series at all (taking for granted any properties of special functions whose derivations require series manipulation):
$$\begin{align*}
I &= \int_0^1 \frac{\log(1-x) \log^n(x)}x \, dx \\[1ex]
&= \int_0^1 \frac{\log(x) \log^n(1-x)}{1-x} \, dx \tag{1} \\[1ex]
&= (-1)^n \int_0^\infty x^n \log\left(1-e^{-x}\right) \, dx \tag{2} \\[1ex]
&= (-1)^{n+1} n \int_0^\infty x^{n-1} \operatorname{Li}_2(e^{-x}) \, dx \tag{3} \\[1ex]
&= (-1)^{n+1} n (n-1) \int_0^\infty x^{n-2} \operatorname{Li}_3(e^{-x}) \, dx = \cdots \tag{4} \\
&\;\vdots \\
&= (-1)^{n+1} n! \int_0^\infty \operatorname{Li}_{n+1}(e^{-x}) \, dx \\[1ex]
&= (-1)^{n+1} n! \operatorname{Li}_{n+2}(1) \\[1ex]
&= \boxed{(-1)^{n+1} \Gamma(n+1) \zeta(n+2)} \tag{5}
\end{align*}$$


*

*$(1)$ : substitute $x\mapsto1-x$

*$(2)$ : substitute $x\mapsto 1-e^{-x}$

*$(3)$ : integrate by parts, recalling $\displaystyle \frac d{dx}\operatorname{Li}_2(x) = -\frac{\log(1-x)}x$ where $\operatorname{Li}_2$ is the dilogarithm

*$(4)$ : integrate by parts ad nauseam, using the recurrence $\displaystyle\frac d{dx}\operatorname{Li}_n(x)=\frac{\operatorname{Li}_{n-1}(x)}x$

*$(5)$ : $n!=\Gamma(n+1)$ and $\operatorname{Li}_n(1)=\zeta(n)$
</s>