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

My Try:: First we will factorise $1+x^8 = 1^2+(x^4)^2+2x^4-2x^4$

$ = (1+x^4)^2-(\sqrt{2}.x^2)^2 = (x^4+\sqrt{2}x^2+1).(x^4-\sqrt{2}x^2+1)$

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

Now for partial fraction solving Let we will take $x^2 = t$

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

But Using This Method solution steps are very Complex

So anyone have a Nice Method to solve Given Question.


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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

1 Answer 1

up vote 6 down vote accepted

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


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.

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