# Integration of $\displaystyle \int\frac{1}{1+x^8}\,dx$

Compute the indefinite integral $$\int\frac{1}{1+x^8}\,dx$$

My Attempt:

First we will factor $1+x^8$

\begin{align} 1+x^8 &= 1^2+(x^4)^2+2x^4-2x^4\\ &= (1+x^4)^2-(\sqrt{2}x^2)^2\\ &= (x^4+\sqrt{2}x^2+1)(x^4-\sqrt{2}x^2+1) \end{align}

Then we can rewrite the integral as

$$\int\frac{1}{1+x^8}\,dx = \int \frac{1}{(x^4+\sqrt{2}x^2+1)(x^4-\sqrt{2}x^2+1)}\,dx$$

To use partial fractions let $t = x^2$ to get

$$\frac {1}{(t^2+\sqrt{2}t+1)(t^2-\sqrt{2}t+1)} = \frac{At+B}{t^2+\sqrt{2}t+1}+\frac{Ct+D}{t^2-\sqrt{2}t+1}$$

This method of solving the problem becomes very complex. Is there a less complex approach to the problem?

-
By the looks of this, I don't think we'll be able to find a nice method to solve this integral. – Jared May 22 '13 at 4:43
@Jared I was curious when there was a step by step solution button xD – DanZimm May 22 '13 at 4:45
What is interesting to me is that there appears to be a nice structure in the integral of $\frac{1}{1+x^n}$ for any $n$ you like. Try, for example, integrals.wolfram.com/… for $n=111$. – Lord Soth May 22 '13 at 4:47
As an aside, $$\int_0^\infty\frac{dx}{1+x^n}=\frac{\frac\pi n}{\sin\left(\frac\pi n\right)}$$ – Lucian Dec 9 '13 at 2:08

Why not splitting up in fractions until you have first degree polynomials in the nominators?

$$\frac{1}{1+x^8}=\frac{A}{x-e^{i\pi/8}}+\frac{B}{x-e^{-i\pi/8}}+\frac{C}{x-e^{i3\pi/8}}+\frac{D}{x-e^{-i3\pi/8}}+\frac{E}{x-e^{i5\pi/8}}+\frac{F}{x-e^{-i5\pi/8}}+\frac{G}{x-e^{i7\pi/8}}+\frac{H}{x-e^{-i7\pi/8}}$$

or if you prefer without the complex numbers

$$\frac{1}{1+x^8}=\frac{ax+b}{x^2-2\cos(\pi/8)x+1}+\frac{cx+d}{x^2-2\cos(3\pi/8)x+1}+\frac{ex+f}{x^2-2\cos(5\pi/8)x+1}+\frac{gx+h}{x^2-2\cos(7\pi/8)x+1} \; .$$

With the complex formula, you can find the coefficients easily as follows

$$A=\lim_{x \to e^{i\pi/8}}\frac{x-e^{i\pi/8}}{1+x^8}\overset{\text{H}}{=}\lim_{x \to e^{i\pi/8}}\frac{1}{8x^7}=\frac{e^{-i7\pi/8}}{8}$$

where I used de l'Hôpital's rule.

-