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

Question

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?

Answer 1

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.

Answer 2

We first decompose the quartic into two quadratics. $$\displaystyle \begin{aligned}\int \frac{1}{1+x^{8}} d x &=\frac{1}{2 \sqrt{2}}\int\frac{1}{x^{2}}\left(\frac{1}{x^{4}-\sqrt{2} x^{2}+1}-\frac{1}{x^{4}+\sqrt{2} x^{2}+1}\right) d x \\ &=\frac{1}{2 \sqrt{2}}\left[\int \frac{d x}{x^{2}\left(x^{4}-\sqrt{2} x^{2}+1\right)}-\int \frac{d x}{x^{2}\left(x^{4}+\sqrt{2} x^{2}+1\right)}\right]\\& =\int\left(\frac{1}{x^{2}}-\frac{x^{2}-\sqrt{2}}{x^{4}-\sqrt{2} x^{2}+1}\right)-\left(\frac{1}{x^{2}}-\frac{x^{2}+\sqrt{2}}{x^{4}+\sqrt{2} x^{2}+1}\right) d x \\&=\int \frac{x^{2}+\sqrt{2}}{x^{4}+\sqrt{2} x^{2}+1} d x-\int \frac{x^{2}-\sqrt{2}}{x^{4}-\sqrt{2} x^{2}+1} d x \\\end{aligned}$$

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For the first integral, we further decompose it into two integrals $J_+$ and $J_-$. $\displaystyle \text {Now let } \quad 1+\frac{\sqrt{2}}{x^{2}} \equiv A\left(1+\frac{1}{x^{2}}\right)-B\left(1-\frac{1}{x^{2}}\right)$, then $ \displaystyle A=\frac{\sqrt{2}+1}{2} \textrm{ and }B=\frac{\sqrt{2}-1}{2}$ $\displaystyle J _ { + } = \frac { \sqrt { 2 } + 1 } { 2 } \int \frac { ( 1 + \frac { 1 } { x ^ { 2 } } ) d x } { x ^ { 2 } + \frac { 1 } { x ^ { 2 } } + \sqrt { 2 } } \\ \displaystyle \quad - \frac { \sqrt { 2 } - 1 } { 2 } \int \frac { ( 1 - \frac { 1 } { x ^ { 2 } } ) d x } { x ^ { 2 } + \frac { 1 } { x ^ { 2 } } + \sqrt { 2 } }$ $\displaystyle \\ \displaystyle \quad = \frac { \sqrt { 2 } + 1 } { 2 } \int \frac { d ( x - \frac { 1 } { x } ) } { ( x - \frac { 1 } { x } ) ^ { 2 } + 2 + \sqrt { 2 } } \\ \quad \displaystyle - \frac { \sqrt { 2 } - 1 } { 2 } \int \frac { d ( x + \frac { 1 } { x } ) } { ( x + \frac { 1 } { x } ) ^ { 2 } + ( 2 - \sqrt { 2 } ) }$ $\displaystyle \quad = \frac { \sqrt { 2 } + 1 } { 2 \sqrt { 2 + \sqrt { 2 } } } \tan ^ { - 1 } \left( \frac { x - \frac { 1 } { x } } { \sqrt { 2 + \sqrt { 2 } } }\right )\\ \displaystyle \quad - \frac { \sqrt { 2 } - 1 } { 4 \sqrt { 2 - \sqrt { 2 } } } \ln \left| \frac { x + \frac { 1 } { x } - \sqrt { 2 - \sqrt { 2 } } } { x + \frac { 1 } { x } + \sqrt { 2 - \sqrt { 2 } } }\right|$ $\displaystyle \quad = \frac { \sqrt { 2 } + 1 } { 2 \sqrt { 2 + \sqrt { 2 } } } \tan ^ { - 1 } ( \frac { x ^ { 2 } - 1 } { \sqrt { 2 + \sqrt { 2 } } x } ) \\ \quad \displaystyle - \frac { \sqrt { 2 } - 1 } { 4 \sqrt { 2 - \sqrt { 2 } } } \ln \left| \frac { x ^ { 2 } - \sqrt { 2 - \sqrt { 2 } } x + 1 } { x ^ { 2 } + \sqrt { 2 - \sqrt { 2 } } x + 1 }\right|$ $\displaystyle \left. \begin{array}{l}{ \quad J_- \displaystyle \\ \displaystyle = \int \frac { x ^ { 2 } - \sqrt { 2 } } { x ^ { 4 } - \sqrt { 2 } x ^ { 2 } + 1 } d x }\\{ \displaystyle = \int \frac { 1 - \frac { \sqrt { 2 } } { x ^ { 2 } } } { x ^ { 2 } + \frac { 1 } { x ^ { 2 } } - \sqrt { 2 } } d x }\end{array} \right.$ $\displaystyle = \int ( \frac { \frac{1 - \sqrt { 2 }}{2} ( 1 + \frac { 1 } { x ^ { 2 } } ) } { x ^ { 2 } + \frac { 1 } { x ^ { 2 } } - \sqrt { 2 } } + \frac { \frac { \sqrt { 2 } + 1 } { 2 } ( 1 - \frac { 1 } { x ^ { 2 } } ) } { x ^ { 2 } + \frac { 1 } { x ^ { 2 } } - \sqrt { 2 } } ) d x$ $\displaystyle =\frac { 1 - \sqrt { 2 } } { 2 } \int \frac { d ( x - \frac { 1 } { x } ) } { ( x - \frac { 1 } { x } ) ^ { 2 } + ( 2 - \sqrt { 2 } ) } \\ \displaystyle \quad + \frac { \sqrt { 2 } + 1 } { 2 } \int \frac { d ( x + \frac { 1 } { x } ) } { ( x + \frac { 1 } { x } ) ^ { 2 } - ( 2 + \sqrt { 2 } )}$ $\displaystyle = - \frac { \sqrt { 2 } - 1 } { 2 \sqrt { 2 - \sqrt { 2 } } } \tan ^ { - 1 }\left ( \frac { x - \frac { 1 } { x } } { \sqrt { 2 - \sqrt { 2 } } } \right) $ $\quad \displaystyle + \frac { \sqrt { 2 } + 1 } { 4 \sqrt { 2 + \sqrt { 2 } } } \ln \left|\frac { x + \frac { 1 } { x } - \sqrt { 2 + \sqrt { 2 } } } { x + \frac { 1 } { x } + \sqrt { 2 + \sqrt { 2 } } }\right | + C$ $=\displaystyle -\frac { \sqrt { 2 } - 1 } { 2 \sqrt { 2 - \sqrt { 2 } } } \tan ^ { - 1 } \left( \frac { x ^ { 2 } - 1 } { \sqrt { 2 - \sqrt { 2 } } x} \right)$ $\displaystyle \quad + \frac { \sqrt { 2 } + 1 } { 4 \sqrt { 2 + \sqrt { 2 } } } \ln \left| \frac { x ^ { 2 } - \sqrt { 2 + \sqrt { 2 } } x + 1 } { x ^ { 2 } + \sqrt { 2 + \sqrt { 2 } } x + 1 }\right | \quad + C$

$$\displaystyle \int \frac{1}{1+x^{8}}dx \\= \frac { \sqrt { 2 } + 1 } { 4 \sqrt { 4 + 2 \sqrt { 2 } } } \tan ^ { - 1 }\left( \frac { x ^ { 2 } - 1 } { \sqrt { 2 + \sqrt { 2 } } x } \right) \quad +\frac { \sqrt { 2 } - 1 } { 8 \sqrt { 4 - 2 \sqrt { 2 } } } \ln \left| \frac { x ^ { 2 } + \sqrt { 2 - \sqrt { 2 } } x + 1 } { x ^ { 2 } - \sqrt { 2 - \sqrt { 2 } x + 1 } } \right| \\ \displaystyle \quad + \frac { \sqrt { 2 } - 1 } { 4 \sqrt { 4 - 2 \sqrt { 2 } } } \tan ^ { - 1 } \left( \frac { x ^ { 2 } - 1 } { \sqrt { 2 - \sqrt { 2 } } x } \right) \quad + \frac { \sqrt { 2 } + 1 } { 8 \sqrt { 4 + 2 \sqrt { 2 } } } \ln \left| \frac { x ^ { 2 }+\sqrt { 2 + \sqrt { 2 }} x + 1 } { x ^ { 2 } - \sqrt { 2 + \sqrt { 2 } } x + 1 } \right| + C $$