How many method to evaluate the integral $\int_{0}^{1} \frac{\ln ^{n}(1-x)}{x} d x , \textrm{ where }n\in N?$ $$
\begin{aligned}
\int_{0}^{1} \frac{\ln (1-x)}{x} d x &\stackrel{x \rightarrow 1-x}{=} \int_{0}^{1} \frac{\ln x}{1-x} d x \\
&=\sum_{k=0}^{\infty} \int_{0}^{1} x^{k} \ln x d x \\
&=\sum_{k=0}^{\infty} \int_{0}^{1} \ln x d\left(\frac{x^{k+1}}{k+1}\right) \\
& \stackrel{I B P}{=} \sum_{k=0}^{\infty}\left(\left[\frac{x^{k+1} \ln x}{k+1}\right]_{0}^{1}-\int_{0}^1{\frac{x^{k}}{k+1}} d x\right) \\
&=-\sum_{k=0}^{\infty} \frac{1}{(k+1)^{2}} \\
&=-\zeta(2)
\end{aligned}
$$
Similarly,
$$
\begin{aligned}
\int_{0}^{1} \frac{\ln ^{2}(1-x)}{x} d x 
\stackrel{x\mapsto 1-x}{=}  & \int_{0}^{1} \frac{\ln ^{2} x}{1-x} d x \\
=& \sum_{k=0}^{\infty} \int_{0}^{1} x^{k} \ln ^{2} x d x \\
\stackrel{IBP}{=} & \sum_{k=0}^{\infty} \frac{1}{k+1}\left(\left[x^{k+1} \ln ^{2} x\right]_{0}^{1}-\int_{0}^{1} 2 x^{k} \ln x d x\right) \\
\stackrel{IBP}{=}&-2 \sum_{k=0}^{\infty} \frac{1}{k+1}\left(\left[\frac{x^{k+1}\ln x}{k+1}\right]_{0}^{1}-\int_{0}^{1} \frac{x^{k}}{k+1} d x\right) \\
=& 2 \sum_{k=0}^{\infty} \frac{1}{(k+1)^{3}} \\
=& 2 \zeta(3)
\end{aligned}
$$
Replacing the power of $\ln x$ by $n$ and performing integration by parts by $n$ times yields $$
\begin{aligned}\int_{0}^{1} \frac{\ln ^{n}(1-x)}{x} d x &\stackrel{x\mapsto 1-x}{=} \int_{0}^{1} \frac{\ln ^{n} x}{1-x} d x\\ &\qquad\qquad \vdots \\&= (-1)(-2)(-3) \cdots(-n) \sum_{k=0}^{\infty} \frac{1}{(k+1)^{n+1}}\\&= (-1)^{n} n ! \zeta (n+1)\end{aligned}
$$
My Question
Is there an alternative method to evaluate the integral?
 A: Right of the top of my head I can see two ways. There might be many more.
1:-
$$\int_{0}^{1} \frac{\ln^{n} (1-x)}{x} dx=\int_{0}^{1}\frac{\ln^{n}(x)}{1-x}\,dx$$.
Using $\sum_{r=0}^{\infty}x^{r}=\frac{1}{1-x}\,\,\,,|x|<1$.
Then Using DCT.
$$\sum_{r=0}^{\infty}\int_{0}^{1}\ln^{n}(x)x^{r}\,dx$$.
Let $x=e^{-t}$
Then :-
$$\sum_{r=0}^{\infty}\int_{0}^{\infty}(-1)^{n}t^{n}e^{-(r+1)t}\,\,dt=$$.
Now substituting $(r+1)t=z$ and using Gamma function you have:-
$$\sum_{r=0}^{\infty}\frac{(-1)^{n}\Gamma(n+1)}{(r+1)^{n+1}}=(-1)^{n}n!\zeta(n+1)$$.
Alternatively:-
$$\int_{0}^{1}(1-x)^{k-1}x^{m-1}\,dx=B(m,k)=\frac{\Gamma(k)\Gamma(m)}{\Gamma(m+k)}$$.
So differentiating partially wrt $m$. You have :-
$$\frac{\partial B(m,k)}{\partial m}=\int_{0}^{1}(1-x)^{k-1}x^{m-1}\ln(x)\,dx$$.
Proceeding inductively:-
$$\frac{\partial^{n} B(m,n)}{\partial m^{n}}\bigg|_{k=0,m=1}=\int_{0}^{1}\frac{\ln^{n}(x)}{1-x}\,dx$$
To differentiate the Gamma function and evaluating , you will need values of the polygamma function.
A: I would like to switch the integration into differentiation process by defining an integral partner $$I(a):=\int_{0}^{1} \frac{\ln(1-x)(1-x)^{a}}{x} d x.$$
Then
$$ \begin{aligned}I(a)& \stackrel{x \mapsto 1-x}{=} \int_{0}^{1} \frac{x^{a}\ln x}{1-x} d x\\&=\sum_{k=0}^{\infty} \int_{0}^{1} x^{k+a} \ln x d x 
\\& \stackrel{IBP}{=} -\sum_{k=0}^{\infty} \frac{1}{(k+a+1)^2}
\end{aligned}$$
Differentiating both sides w.r.t. $a$ by $n-1$ times yields $$
\begin{aligned} \int_{0}^{1} \frac{\ln (1-x)}{x} \frac{\partial^{n-1}}{\partial a^{n-1}}(1-x)^{a} d x&=-(-2)(-3)\cdots(-n) \sum_{k=0}^{\infty} \frac{1}{(k+a+1)^{n+1}}\\\int_{0}^{1} \frac{1}{x}(1-x)^{a} \ln ^{n}(1-x) dx&= (-1)^nn!\sum_{k=0}^{\infty} \frac{1}{(k+a+1)^{n+1}}
\end{aligned}
$$
Now we can conclude that
\begin{aligned}
\int_{0}^{1} \frac{\ln ^{n}(1-x)}{x} d x &=I^{(n)}(0)\\ &=(-1)^{n} n ! \sum_{k=0}^{\infty} \frac{1}{(k+1)^{n+1}} \\
&=(-1)^{n} n !\zeta(n+1)
\end{aligned}
