Methods to derive a closed form for $I_n=\int_0^\infty \frac{\ln^n(x+1)-\ln^n(x)}{x+1}dx$

I've stumbled onto this general integral that has closed form values for the $$n\in \Bbb{Z^+}$$ $$I_n=\int_0^\infty \frac{\ln^n(x+1)-\ln^n(x)}{x+1}dx$$

Obviously $$I_0=0$$ but higher values of $$n$$ yield interesting results. The first few are: $$I_1=\frac{\pi^2}{6}$$ $$I_2=0$$ $$I_3=\frac{7\pi^4}{60}$$ $$I_4=0$$ $$...$$ So it appears that $$I_{2n}=0$$ and that $$I_{2n+1}=C_{2n+1}\zeta\left(2(n+1)\right)$$ where $$C_{n}$$ is some rational number that is equal to $$0$$ when $$n\in2\Bbb{Z^+}$$. Can anyone explain why $$I_{2n}=0$$ and how one can derive $$I_{2n+1}$$?

$$I_n=\int_0^\infty\frac {ln^n(x+1)-ln^n(x)}{x+1}dx=\int_1^\infty\frac{ln^n(x)}{x(x+1)}dx-\int_0^1\frac{ln^n(x)}{1+x}dx$$, using $$x+1 \to x$$ in the $$ln^n(x+1)$$ term.
Let $$y=\frac{1}{x}$$ in the first integral and it becomes $$\int_0^1\frac{(-ln(y))^n}{1+y}dy$$.
Net result $$I_n=0$$ for even $$n$$ and $$I_n=-2\int_0^1\frac{ln^n(x)}{1+x}dx$$ for odd $$n$$. This last integral can be found in Gradshteyn and Ryzhik Table of Integrals...
• From the Table I referenced: $\int_0^1\frac{(lnx)^{2n-1}}{1+x}dx=\frac{1-2^{2n-1}}{2n}\pi^{2n}|B_{2n}|$ where $B_{2n}$ are Bernoulli numbers. If you can access that table, look for section 4.271. $– herb steinberg May 2 at 19:38 • I see, I also realized that the integral you had$I_n\$ reduced to can be evaluated by a simple substitution to yield the answer in terms of the Gamma and Dirichlet Eta Function. – aleden May 2 at 19:40