how to compute $\int_{0}^{\infty} \frac{dx}{x^{2017}+1}$ How do I simplify this following integral
$$\int_{0}^{\infty} \frac{dx}{x^{2017}+1}$$
I think the answer is
 $\frac{1}{2017}\pi \csc\left(\frac{\pi}{2017}\right)$
which is really close to one, what is the best method to solve this ?  I was trying compute it with residue integral but I noticed this integral does have a massive poles. Any ideas?

To expand on my comment, the given integral is equal to $$\int_{0}^{\infty}\frac{dx}{x^{n}+1}=\frac{1}{n}\int_{0}^{\infty}\frac{t^{1/n-1}}{1+t}\,dt=\frac{B(1/n,1-1/n)}{n}=\frac{\Gamma(1/n)\Gamma(1-1/n)}{n\Gamma(1)}=\frac{\pi}{n\sin(\pi/n)}$$ In the above we have used the substitution $x^{n} =t$ and the properties of Gamma and Beta functions namely $$B(m, n) =\int_{0}^{\infty}\frac{x^{m-1}}{(1+x)^{m+n}}\,dx=\frac{\Gamma (m) \Gamma (n)} {\Gamma (m+n)}, \, \Gamma (x) \Gamma (1-x)=\frac{\pi}{\sin x\pi} $$
