How can we evaluate the integral $\int_{0}^{\infty} \frac{d x}{1+x+x^{2}+\ldots+x^{n-1}}$, where $n\geq 3$? We are going to investigate the integral
$$
I_{n}=\int_{0}^{\infty} \frac{d x}{1+x+x^{2}+\ldots+x^{n-1}} \text {, where } n \geqslant 3.
$$
Let’s start with the simpler cases.
$$
\begin{aligned}
I_{3} &=\int_{0}^{\infty} \frac{d x}{1+x+x^{2}} \\
&=\int_{0}^{\infty} \frac{d x}{\left(x+\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}} \\
&=\frac{2}{\sqrt{3}}\left[\tan ^{-1}\left(\frac{2 x+1}{\sqrt{3}}\right)\right]_{0}^{\infty} \\
&=\frac{2}{\sqrt{3}}\left(\frac{\pi}{2}-\frac{\pi}{6}\right) \\
&=\frac{2 \pi}{3 \sqrt{3}}
\end{aligned}
$$
and
$$
\begin{aligned}
I_{4} &=\int_{0}^{\infty} \frac{1}{(1+x)\left(1+x^{2}\right)} d x \\
&=\frac{1}{2} \int_{0}^{\infty}\left(\frac{1}{x+1}+\frac{1-x}{x^{2}+1}\right) d x \\
&=\frac{1}{2}\left[\ln (x+1)+\tan ^{-1} x-\frac{1}{2} \ln \left(x^{2}+1\right)\right]_{0}^{\infty} \\
&=\frac{1}{4}\left[\ln \frac{(x+1)^{2}}{x^{2}+1}\right]_{0}^{\infty}+ \left[\frac{1}{2} \tan ^{-1} x\right]_{0}^{\infty} \\
&=\frac{\pi}{4}
\end{aligned}
$$
But the integrals are difficult when $n\geq 5$.
My question: Is there any elementary method to evaluate it?
Your suggestions and solutions are warmly welcome.

You must have an error somewhere in the indices because
$$I_n=\int_0^\infty \frac{x-1}{x^{n+1}-1}\,dx = \frac{\pi  }{n+1}\csc \left(\frac{2 \pi }{n+1}\right)$$
There is another (but not elementary) method to find another form of the result using hypergeometric functions.
$$J_n=\int\frac{x-1}{x^{n+1}-1}\,dx =x \, _2F_1\left(1,\frac{1}{n+1};\frac{n+2}{n+1};x^{n+1}\right)-$$ $$\frac{1}{2} x^2 \,
   _2F_1\left(1,\frac{2}{n+1};\frac{n+3}{n+1};x^{n+1}\right)$$ which leads to
$$I_n=\cos \left(\frac{\pi }{n+1}\right) \Gamma \left(\frac{n}{n+1}\right) \Gamma
   \left(\frac{n+2}{n+1}\right)-$$ $$\frac{1}{2} \cos \left(\frac{2 \pi }{n+1}\right)
   \Gamma \left(\frac{n-1}{n+1}\right) \Gamma \left(\frac{n+3}{n+1}\right)$$
