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?

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.
