# Does $\int_0^\infty\frac{\ln(1+x)}{x(1+x^n)}dx$ have a general form?

Does $$I_n=\int_0^\infty\frac{\ln(1+x)}{x(1+x^n)}dx$$ have a general form?

I tried to evaluate some small $$n$$s.
For $$n=1$$, $$I_1$$ is obviously $$\frac16\pi^2$$.
For $$n=2$$, see here. $$I_2=\frac5{48}\pi^2$$.
For $$n=3$$, I put it in Mathematica and get $$\small{\frac{1}{108} \left(9 \left(4 \left(\text{Li}_2\left(\frac{\sqrt[6]{-1}}{\sqrt{3}}\right)+\text{Li}_2\left(-\frac{(-1)^{5/6}}{\sqrt{3}}\right)\right)+\log ^2(3)\right)+5 \pi ^2\right)}$$ Use the result $$\Re\operatorname{Li}_2\left(\frac{1+ti}2\right)=\frac1{12}\pi^2-\frac12\arctan^2t-\frac18\ln^2\frac{1+t^2}4,$$ I'm able to show $$I_3=\frac5{54}\pi^2$$.
For $$n=4$$, I numerically found $$I_4=\frac{17}{192}\pi^2$$.
I'm not able to find the general form with $$n\in \mathbb{Z}^+$$.

• Sorry for not perceiving the answer in the linked question can be applied in this question. Feel free to close this question as a duplicate of that one. Dec 13, 2018 at 2:02

\begin{aligned} I_n &= \int_0^1 \frac{\ln(1+x)}{x(1+x^n)}dx + \int_1^\infty \frac{\ln(1+x)}{x(1+x^n)}dx \\ &= \int_0^1 \frac{\ln(1+x)}{x(1+x^n)}dx + \int_0^1 \frac{x^n \ln(1+x)}{x(1+x^n)}dx - \int_0^1 \frac{x^{n-1} \ln x}{1+x^n}dx \\ &= \int_0^1 \frac{\ln(1+x)}{x}dx + \frac{1}{n}\int_0^1 \frac{\ln(1+x^n)}{x}dx \\ &= \int_0^1 \frac{\ln(1+x)}{x}dx+\frac{1}{n^2}\int_0^1 \frac{\ln(1+u)}{u}du = \color{blue}{(1+\frac{1}{n^2})\frac{\pi^2}{12}} \end{aligned}
• Second line: $$x\mapsto 1/x$$
• Fourth line: $$x=u^{1/n}$$
Note that $$n$$ need not be an integer.