Since the original question is heavily downvoted, I'm not sure if the moderators will delete it before I can post my answer.
Here's my attempt:
Following JG's remarks, we have
$$ x^5 + 1 = \frac14(x+1)(2x^2 + x(\sqrt5 - 1) + 2)(2x^2 - x(\sqrt5 - 1) + 2) $$
Then by partial fractions, we have $$ \dfrac1{x^5 + 1} = \dfrac4{(x+1)(2x^2 + x(\sqrt5 - 1) + 2)(2x^2 - x(\sqrt5 - 1) + 2)} $$ is equal to $$ \dfrac2{5+\sqrt5} \cdot \dfrac1{x+1} -\dfrac{4\sqrt5}{(\sqrt5 - 1)(5+\sqrt5)} \cdot \dfrac x{2x^2 - x(\sqrt5 - 1) + 2} + \dfrac4{(\sqrt5-1)(5+\sqrt5)} \cdot \dfrac{x + \sqrt5 + 1}{2x^2 + x(\sqrt5 - 1) + 2}$$
What's left is to apply $ \int \dfrac1{x^2+a^2} \, dx = \frac1a \tan^{-1} (\frac xa) + C$ and $ \int \dfrac{1}{x^2-a^2}\, dx = -\frac 1a \tanh^{-1} (\frac xa) + C$.
Obviously all the calculations above are fairly tedious. Is there a(another) way to evaluate this indefinite integral?
To clarify, I'm not interested to see ANY roots of unity ($\omega $) in the final result.
Naturally, I'm also interested if there's other approach is applicable for the indefinite integral of $\frac1{x^n + 1}$ for all positive integers $n$.
First, integrate $g(x,\phi)=\frac{2\phi x-2}{x^2-2\phi x+1}$ \begin{align} I(x,\phi) &= \int g(x,\phi)dx =\phi\ln\left(x^2-2\phi x+1\right) -2\sqrt{1-\phi^2} \tan^{-1}\frac{x-\phi}{\sqrt{1-\phi^2}} \end{align} Then, with $\phi_{\pm} = \frac{1\pm\sqrt5}{4}$ \begin{align} \int \frac{1}{1+x^5}dx &=\frac15 \int \left(\frac1{x+1}- g(x,\phi_+) - g(x,\phi_-) \right) dx\\ &=\frac15\left[\ln(x+1)-I(x,\phi_+)-I(x,\phi_-)\right] + C \end{align}