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How do I find $\displaystyle\int\dfrac{dx}{1+x^8}$?

My friend asked me to find $\displaystyle\int\dfrac{dx}{1+x^{2n}}$ for a positive integer $n$. But looking up I am getting pretty noisy answer for a general value.

I have seen that $\displaystyle\int\dfrac{dx}{1+x^6}$ can be broken into partial fractions because of the odd factor of $6$. So I am curious what is the algorithm to compute the integral for $n$ being a power of $2$.

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  • $\begingroup$ Is it a definie or indefinite integral you are looking for? looking at wolframalpha the indefinite looks really horrible wolframAlpha $\endgroup$ – Noxet Jul 23 '14 at 12:57
  • $\begingroup$ Indefinite and I have seen the wolframalpha output. $\endgroup$ – Grobber Jul 23 '14 at 12:58
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    $\begingroup$ Let $\omega_1, \omega_2, \ldots, \omega_n$ be the set of roots of $x^n+1$, since all the roots are simple $$\frac{1}{x^n+1} = \sum_{k=1}^n\frac{1}{n\omega_k^{n-1}(x-\omega_i)} = -\frac{1}{n}\sum_{k=1}^n \frac{\omega_k}{x-\omega_k}$$ $\endgroup$ – achille hui Jul 23 '14 at 13:07
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    $\begingroup$ Related: Solving this integral? See also the .pdf files I posted here, which give full details for (U.S.) calculus 2 methods of evaluating the antiderivatives of $\frac{1}{1+x^n}$ for $n=4,$ $5,$ and $6.$ In the case of $n=7,$ there isn't an answer similar to those I gave that makes use of numbers expressible in terms of radicals of real numbers, but it is possible for $n=8,$ and I think the difficulty for $n=8$ will probably be between that for $n=4$ and $n=5.$ $\endgroup$ – Dave L. Renfro Jul 23 '14 at 13:34
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$$x^8+1=x^8+2x^4+1-2x^4=(x^4+1)^2-2x^4=(x^4+\sqrt{2}x^2+1)(x^4-\sqrt{2}x^2+1)$$

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

Therefore $$x^8+1=(x^2+1 -\sqrt{2 + \sqrt{2}}x)(x^2+1 +\sqrt{2 + \sqrt{2}}x)(x^2+1 -\sqrt{2 - \sqrt{2}}x)(x^2+1 +\sqrt{2 - \sqrt{2}}x)$$

Now the partial fraction decomposition is ugly but doable.

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