<s>[INST] Elementary solution to $ \int \frac{1}{x^5+1} \, dx $ 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 $n$.
 [/INST] Let $g(x,\phi)=\frac{2\phi x-2}{x^2-2\phi x+1}$ and integrate
\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}
</s>