# 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.

• There is no simple way. See here the solution, and convince yourself that your goal is hopeless. – Crostul Nov 14 '16 at 23:28
• Before posting such questions, at the minimum, you should check Wolfram Alpha: wolframalpha.com/input/?i=int+1%2F(1%2Bx%5E5) – poweierstrass Nov 14 '16 at 23:29

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.

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

$$x_{kr}=\cos \left(\frac{(2k-1)\pi}{n}\right)$$

$$x_{ki}=\sin \left(\frac{(2k-1)\pi}{n}\right)$$

$$\int\frac1{1+x^5}dx=-\frac15\sum_{k=1}^5\left(\frac12x_{kr}\log(x^2-2x_{kr}x+1)-x_{ki}\arctan\left(\frac{x-x_{kr}}{x_{ki}}\right)\right)+C$$
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.