Evaluate: $\int_{0}^{\infty}\left[\frac{\ln(1+x)}{1+x}\right]^n\mathrm dx$

Question

I am looking for the closed form, but I can't...

$$\int_{0}^{\infty}\left[\frac{\ln(1+x)}{1+x}\right]^n\mathrm dx=G(n)$$

$n\ge2$

We have: $G(2)=2$, $G(3)=\frac{3}{8}$,$G(4)=\frac{8}{81}$, $G(5)=\frac{15}{512}$, and so on,...

$$\int_{0}^{\infty}u^n e^{u(1-n)}\mathrm du\tag1$$

Is there an easy of calculating $(1)$

Answer 1

$$ \int_{0}^{\infty}u^n e^{u(1-n)}\mathrm du = \frac{1}{(n-1)^{n+1}} \int_0^\infty e^{-t} t^{n} \;dt = \frac{\Gamma(n+1)}{(n-1)^{n+1}} = \frac{n!}{(n-1)^{n+1}} $$

Answer 2

Starting from $$G(n) = \int_0^\infty u^n e^{u(1-n)}du$$

One has, by letting $(n-1)u=x \implies (n-1) du =dx$, then \begin{align} G &= \int_0^\infty\left( \frac{x}{n-1}\right)^ne^{-x}dx \\ &= \frac{\Gamma(n+1)}{(n-1)^{n+1}} \end{align}