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 $n\in \mathbb{Z}^+$.

$$\begin{aligned}
I_n &= \int_0^1 \frac{\ln(1+x)}{x(1+x^n)}dx + \int_1^\infty \frac{\ln(1+x)}{x(1+x^n)}dx \\
&= \int_0^1 \frac{\ln(1+x)}{x(1+x^n)}dx + \int_0^1 \frac{x^n \ln(1+x)}{x(1+x^n)}dx - \int_0^1 \frac{x^{n-1} \ln x}{1+x^n}dx \\
&= \int_0^1 \frac{\ln(1+x)}{x}dx + \frac{1}{n}\int_0^1 \frac{\ln(1+x^n)}{x}dx \\
&= \int_0^1 \frac{\ln(1+x)}{x}dx+\frac{1}{n^2}\int_0^1 \frac{\ln(1+u)}{u}du = \color{blue}{(1+\frac{1}{n^2})\frac{\pi^2}{12}}
\end{aligned}$$


*

*Second line: $x\mapsto 1/x$ 

*Third line: Integration by parts 

*Fourth line: $x=u^{1/n}$
Note that $n$ need not be an integer.
