Evaluating $\int_{0}^{1} \frac{\ln^{n} x}{(1-x)^{m}} \, \mathrm dx$

Substitute $x=e^{-t}$ and get that the integral is equal to

$$(-1)^n \int_0^{\infty} dt \, e^{-t} \frac{t^n}{(1-e^{-t})^m} $$

Now use the expansion

$$(1-y)^{-m} = \sum_{k=0}^{\infty} \binom{m+k-1}{k} y^k$$

and reverse the order of summation and integration to get

$$\sum_{k=0}^{\infty} \binom{m+k-1}{k} \int_0^{\infty} dt \, t^n \, e^{-(k+1) t}$$

I then get as the value of the integral:

$$\int_0^1 dx \, \frac{\ln^n{x}}{(1-x)^m} = (-1)^n\, n!\, \sum_{k=0}^{\infty} \binom{m+k-1}{k} \frac{1}{(k+1)^{n+1}}$$

Note that when $m=0$, the sum reduces to $1$; every term in the sum save that at $k=0$ is zero.

Note also that this sum gives you the ability to express the integral in terms of a Riemann zeta function for various values of $m$, which will provide the Stirling coefficients.

Here is another approach based on the beta function

$$ F = \int_{0}^{1}x^{\alpha}(1-x)^{-m} dx = \beta( \alpha+1,1-m ),\quad \alpha+1>0,\, 1-m>0. $$

Now, our integral $I$ can be evaluated as

$$ I = \lim_{\alpha \to 0} \frac{d^n}{d \alpha^n} \beta( \alpha+1,1-m ),\quad n>m-1,\,m\geq 2. $$

Here is a useful identity

$$ {\partial \over \partial x} \mathrm{B}(x, y) = \mathrm{B}(x, y) \left( {\Gamma'(x) \over \Gamma(x)} - {\Gamma'(x + y) \over \Gamma(x + y)} \right) = \mathrm{B}(x, y) (\psi(x) - \psi(x + y)), $$

where $\psi(x)$ is the digamma function.