Closed-form for rational log integral: $\int_0^1\left(\frac{\ln x}{1-x}\right)^{n}dx$ If I may, I have a rather challenging integral.  I am not so sure there is a closed form. 
$$\int_0^1 \left(\frac{\ln x}{1-x}\right)^n \; dx$$
I have evaluated when $n=1$ and $2$. But, when $n=3, 4, 5,\ldots$,  the solution involves varying zeta functions.
For instance, If $n=3$, we get 
$$\begin{align*}
&-3\zeta(3)-\frac{\pi^2}{2}\\
n=4: &\;\ 12\zeta(3)+\frac{2\pi^2}{3}+\frac{4\pi^4}{45}\\
n=5: &\;\ -30\zeta(5)-30\zeta(3)-\frac{11\pi^4}{18}-\frac{5\pi^2}{6}
\end{align*}$$
I have evaluated $$\int_0^1 \frac{\ln^n(x)}{1-x} \; dx=(-1)^n n!\zeta(n+1),$$ but that $n$ power in the denominator of the problem at hand changes things drastically. 
Do you think it is possible to find a closed form? As I said, it may not have one but it would interesting to see some clever methods incorporating the zeta function.
By the way, Maple gave me those values for $n=3,4,5$.
 A: I evaluated the integral you mentioned above and arrived at $$\int_{0}^{1}t^{m}ln^{n}(t)dt=(-1)^{n}\frac{n!}{(m+1)^{n+1}}$$.
May I ask how this is used to evaluate the problem at hand?. I tried using $(1-t)^{m}$ instead of $t^{n}$ letting $m=-n$. But it turned nasty when I used parts as I done with the first part. I am certain there is something I am not seeing.  I did manage to get the $(-1)^{n}\cdot n$ portion as in RobJohn's solution, but not the Stirling/zeta portion 
Here's how I managed the integral you mentioned.  I just do not know how to relate it to the one I posted.  
$$\int_{0}^{1}t^{m}ln^{n}(t)dt$$
Use parts and get:
$$\frac{t^{m+1}}{m+1}ln^{n}(t)-\frac{n}{m+1}\int_{0}^{1}t^{m}ln^{n}(t)dt$$
Using the limits gives:
$$\frac{-n}{m+1}\int_{0}^{1}t^{m}ln^{n-1}(t)dt$$........[1]
Change n to n-1:
$$\int_{0}^{1}t^{m}ln^{n-1}(t)dt=-\frac{n-1}{m+1}\int_{0}^{1}t^{m}ln^{n-2}(t)dt$$
Sub this into [1]:
$$(-1)^{2}\frac{n(n-1)}{(m+1)^{2}}\int_{0}^{1}t^{m}ln^{n-2}(t)dt$$
Now, continue repeating and generally we have:
$$\int_{0}^{1}t^{m}ln^{n}(t)dt=(-1)^{n}\frac{n!}{(m+1)^{n}}\int_{0}^{1}x^{m}dx$$
$$=(-1)^{n}\frac{n!}{(m+1)^{n+1}}$$
Now, how can I relate to the Stirling and zeta to arrive at the closed form RobJohn showed?. 
I tried the same sort of method and let $m=-n$.  This kind of threw a wrench in the whole mess.  I managed to see the $(-1)^{n}\cdot n$, but not the stirling/zeta portion.  
Using parts: $$dv=(1-t)^{-n}dt, \;\ u=ln^{n}(t)dt, \;\ du=\frac{nln^{n-1}(t)}{t}dt$$
$$v=\frac{1}{(1-t)^{n-1}(n-1)}$$
This leads to $$-\frac{n}{n-1}\int_{0}^{1}\frac{ln^{n-1}(t)}{t(1-t)^{n-1}}dt$$
That extra t in the denominator may be a culprit. Otherwise, it is $I_{n-1}$.  
I could repeat as before, but sorry to say I got hung up.
Thanks for your input and help. 
A: In this answer I show that
$$
\int_0^1\left(\frac{\log(t)}{1-t}\right)^n\mathrm{d}t=(-1)^nn\sum_{j=0}^{n-1}\genfrac{[}{]}{0}{0}{n-1}{j}\zeta(n-j+1)
$$
where $\genfrac{[}{]}{0}{0}{n}{k}$ is a Stirling number of the first kind.
A: Well, you can try
$$
A_{n,m} = \int_0^1 (\log x)^n t^m\,dt
$$
which has a nice closed form, then expand yours in terms of that.
