Evaluate the integral $\int \frac{dx}{1+x^5}$ 
$$\int \frac{dx}{1+x^5}$$

I have tried to add a $x^5$ and subtract $x^5$, but got nothing.
 A: Hint. One may start with
$$
x^5+1=(x+1)\left(x^2-\frac{\sqrt{5}+1}{2} x+1\right)\left(x^2+\frac{\sqrt{5}-1}{2} x+1\right)
$$ then one may obtain a partial fraction decomposition,
$$
\frac1{x^5+1}=\frac{a_0}{x+1}+\frac{a_1x+b_1}{x^2-\frac{\sqrt{5}+1}{2} x+1}+\frac{a_2x+b_2}{x^2+\frac{\sqrt{5}-1}{2} x+1}
$$ and integrate classically each term.
A: In this question, I posed the more general problem and found a fantastic answer thanks to Dr. MV.
$$\int\frac1{1+x^n}dx=-\frac1n\sum_{k=1}^n\left(\frac12 x_{kr}\log(x^2-2x_{kr}x+1)-x_{ki}\arctan\left(\frac{x-x_{kr}}{x_{ki}}\right)\right)+C$$
where we have
$$x_{kr}=\cos \left(\frac{(2k-1)\pi}{n}\right)$$
$$x_{ki}=\sin \left(\frac{(2k-1)\pi}{n}\right)$$
For your problem,
$$\int\frac1{1+x^5}dx=-\frac15\sum_{k=1}^5\left(\frac12x_{kr}\log(x^2-2x_{kr}x+1)-x_{ki}\arctan\left(\frac{x-x_{kr}}{x_{ki}}\right)\right)+C$$
A: Actually, this integral is not fundamentally difficult, although as is obvious from the final answer, the algerbra involved is farily "bulky".
The point is that $1+x^5$ factors into five complex roots whose imaginary and real components are related to the sine and cosine of multiples of $\pi/5$.  These are simple algebraic quantities; for example, 
$$
\cos (\pi/5) = \frac{1+\sqrt{5}}{4} \\
\sin(\pi/r) = \sqrt{ \frac58 - \frac{\sqrt{5}}{8}}
$$
so one root is 
$$
\frac{1+\sqrt{5}}{4} + i \sqrt{ \frac58 - \frac{\sqrt{5}}{8}}
$$
Given the five roots, it is easy to decompose into partial fractions, and each of those has straightforward integrals.
