Prove that $\int_0^\infty \frac{\ln x}{x^n-1}\,dx = \Bigl(\frac{\pi}{n\sin(\frac{\pi}{n})}\Bigr)^2$ This question inspired me to ask the following.
Prove that
$$I_n = \int_0^\infty \frac{\ln x}{x^n-1}\,dx = \left(\frac{\pi}{n\sin\left(\frac{\pi}{n}\right)}\right)^2,$$
for $\Re(n)>1$.
For some cases there is a nice specific form of $I_n$. For example
$$\begin{align}
I_2 & = \frac{\pi^2}{4} \\
I_3 & = \frac{4\pi^2}{27} \\
I_4 & = \frac{\pi^2}{8} \\
I_5 & = \frac{2\left(5+\sqrt 5\right)\pi^2}{125} \\
I_6 & = \frac{\pi^2}{9} \\
I_7 & = \frac{2\pi^2}{49\left(1-\sin\left(\frac{3\pi}{14}\right)\right)} \\
I_8 & = \frac{\left(2+\sqrt 2\right)\pi^2}{32}
\end{align}$$

\begin{align}
\int_0^\infty \frac{\ln x}{x^n-1}\,dx
 =& \>\frac1n\int_0^\infty  \int_0^\infty \frac{1}{(1+y)(x^n+y)} dy \ dx \\
=&\>\frac{\pi }{n^2}\csc\frac\pi n \int_0^\infty\frac{y^{\frac1n-1}}{1+y}dy
=\frac{\pi^2}{n^2}\csc^2\frac\pi n
\end{align}
