Given this integral
$$\int_{0}^{\infty}\ln^k\left({1+{1\over x}}\right) \cdot{\mathrm dx\over (1+x)^n}=F(k,n)\tag1$$
Where k and n are integers, $n\ge2$ and $k\ge0$
Setting $n=2$, we noticed that
$$\int_{0}^{\infty}\ln^k\left({1+{1\over x}}\right) \cdot{\mathrm dx\over (1+x)^2}=k!\tag2$$
We noticed that for $n\ge2$ then $F(k,n)$ is conjecture to be rational, how can we show that?
An attempt:
Perform the substitution $t=x/(x+1)$. The integral then reads $$F(k,n)= (-1)^{k} \int_{0}^1\!dt\, (1-t)^{n-2}\log^k t.$$
Now, we use the fact that $$\int_0^1\!dt\,t^j\,\log^k t = \frac{(-1)^k k!}{(j+1)^{k+1}}.\tag{1}$$
We obtain $$F(k,n)= (-1)^{k} \int_{0}^1\!dt\, (1-t)^{n-2}\log^k t = \sum_{j=0}^{n-2} \binom{n-2}{j} (-1)^{k} \int_0^1\!dt\,(-1)^j t^j\,\log^k t =\sum_{j=0}^{n-2} \binom{n-2}{j}\frac{(-1)^j k!}{(j+1)^{k+1}} $$ which proves your conjecture.
What is left to show is the result (1). For that you can observe that $$\log^k t= \lim_{\alpha\to0}\frac{d^k}{d\alpha^k} t^\alpha.$$ Using that we obtain $$\int_0^1\!dt\,t^j\,\log^k t = \lim_{\alpha\to0}\frac{d^k}{d\alpha^k} \int_0^1\!dt\,t^{j+\alpha} =\lim_{\alpha\to0}\frac{d^k}{d\alpha^k} \frac{1}{j+1+\alpha}= (-1)^k \lim_{\alpha\to0} \frac{k!}{(j+1+\alpha)^{k+1}}$$ and the result follows.