Closed form for $\int_0^1 \frac {\log^n(x)}{(1-x)^m} dx$ How can one find a general form for  $\int_0^1 \frac {\log(x)}{(1-x)} dx=-\zeta(2)
\,?$ Namely $\int_0^1 \frac {\log^n(x)}{(1-x)^m} dx\,$ where $n,m\ge1$  Similar to the original  integral I let $1-x=u\,$ which gives $$\int_{-1}^0 \frac {\log^n(1+x)}{x^m} dx$$ and expanding into series we have: $\int_{-1}^0x^{-m}(\sum_{k=1}^{\infty}\frac{(-1)^{k+1}x^k}{k})^n\,dx$ Now this might be doable with a computer using Cauchy product's but otherwise it's  a madness.
Another try is to let $I(k)=\int_0^1 \frac {x^k}{(1-x)^m}\,dx$ And take derivate n times while assuming $k\ge n$ so: $$\frac{d^n}{dx^n}I(k)=\int_0^1\frac{x^k\log^n(x)}{(1-x)^m}dx$$ Plugging $(1-x)^{-m}=\sum_{j=0}^{\infty} \binom{-m}{j}(-1)^jx^j $ in integral and make use of Tonelli
s theorem we get: $$\frac{d^n}{dx^n}I(k)=\sum_{j=0}^{\infty} \binom{-m}{j}(-1)^j\int_0^1 x^{(k+j)}\log^n(x)dx=\sum_{j=0}^{\infty} \binom{-m}{j}(-1)^{(n+j)} n! (k+j+1)^{-(n+1)}$$ But I don't know how to evaluate the latter series.
 A: $$\int_{0}^{1} x^s (1-x)^{-m}\,dx =B(s+1,1-m)=\frac{\Gamma(s+1)\Gamma(1-m)}{\Gamma(s+2-m)}$$
and both sides can be differentiated with respect to $s$ multiple times, then evaluated at $s\to 0^+$.
For differentiating the RHS it is practical to exploit $f'(z)=f(x)\cdot\frac{d}{dz}\log f(z)$ and the fact that $\psi(x)=\frac{d}{dx}\log\Gamma(x)$ fulfills
$$ \psi'(a)=\sum_{n\geq 0}\frac{1}{(n+a)^2} $$
hence $\int_{0}^{1}\frac{\log(x)^n}{(1-x)^m}\,dx$ is naturally related to the values of $\zeta(s)$ for $s\in\{2,3,4,\ldots\}$.
A: 
Be careful: for convenience of the following derivation I have changed $m$ to $m+1$.

We are going to prove that for all integer $n>m\ge0$:
$$
S(n,m):=\int_0^1\frac{\log^n(1-u)}{u^{m+1}}du=\frac{(-1)^n n!}{m!}\sum_{i=0}^{m}{m \brack i}\zeta(n+1-i).\tag{1}
$$
where ${m \brack i}$ are the Stirling numbers of the first kind and $\zeta(n)$ are the Riemann functions.
First we check that the expression is valid for $m=0$ and arbitrary $n>0$:
$$\begin{align}
(-1)^nS(n,0)&=(-1)^n\int_0^1\frac{\log^n(1-u)}{u}du\\
&\stackrel{1-u\mapsto e^{-t}}{=}\int_0^{\infty}\frac{t^n e^{-t}}{1-e^{-t}}dt\\
&=\int_0^{\infty} t^n\sum_{k=1}^\infty e^{-kt}\; dt\\
&=\sum_{k=1}^\infty\int_0^{\infty} t^n e^{-kt}\; dt\\
&\stackrel{t\mapsto z/k}{=}
\sum_{k=1}^\infty\frac{1}{k^{n+1}} \int_0^{\infty}z^n e^{-z}\; dz\\
&=n!\zeta(n+1).
\end{align}$$
Assume now that (1) is valid for some $m\ge0$ and arbitrary $n> m$. We will show that this implies that the expression is valid for $m+1$ and arbitrary $n> m+1$.
$$\begin{align}
S(n,m)&=\int_0^1\frac{\log^{n}(1-u)}{u^{m+1}}du\\
&=-\frac{1}{n+1}\underbrace{\left[\frac{(1-u)\log^{n+1}(1-u)}{u^{m+1}}\right]_0^1}_{=0}\\
&\quad\quad+\frac{1}{n+1}\int_0^1\left(\frac{m}{u^{m+1}}-\frac{m+1}{u^{m+2}}\right)\log^{n+1}(1-u)du\\
&=\frac{m}{n+1}S(n+1,m)-\frac{m+1}{n+1}S(n+1,m+1)
\end{align}$$
or
$$\begin{align}
S(n+1,m+1)&=\frac{m}{m+1}S(n+1,m)-\frac{n+1}{m+1}S(n,m)\\
&\stackrel{I.H.}{=}\frac{m}{m+1}\frac{(-1)^{n+1}(n+1)!}{m!}\sum_{i=0}^{m}{m \brack i}\zeta(n+2-i)\\
&\quad\quad-\frac{n+1}{m+1}\frac{(-1)^n n!}{m!}\sum_{i=0}^{m}{m \brack i}\zeta(n+1-i)\\
&=\frac{(-1)^{n+1}(n+1)!}{(m+1)!}\left[\sum_{i=0}^{m}m{m \brack i}\zeta(n+2-i)+\sum_{i=1}^{m+1}{m \brack i-1}\zeta(n+2-i)\right]\\
&\stackrel{*}{=}\frac{(-1)^{n+1}(n+1)!}{(m+1)!}\sum_{i=0}^{m+1}{m+1 \brack i}\zeta(n+2-i),
\end{align}$$
where in ($\stackrel{*}{=}$) the well-known recurrence identity:
$$
m{m \brack i}+{m \brack i-1}={m+1 \brack i}
$$
was used.
Thus, by induction the claim $(1)$ is proved.

Note added:
If one considers formally the case of "negative" $m$ an interesting kind of symmetry can be observed:

$$
\int_0^1u^m\log^n(1-u)\;du=(-1)^n n!\sum_{i=0}^{m}\binom{m}{i}\frac{(-1)^i}{(i+1)^{n+1}}.
$$

