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Evaluate: $$\int{\frac{x^{5}-x}{x^{8}+1}\:\mathrm dx}.$$

I am unable to see a decent starting point for this integral, there are no radicals so trigonometric substitution isn't helpful; there is no nice partial fraction decomposition to simplify the integrand, integration by parts doesn't help to simplify it much, and I cannot see any factorization or useful substitution to use.

Can anyone help shed some light on this integral?

Thanks in advance!

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    $\begingroup$ When you said 'nice' partial fraction decomposition, does that disbar $x^8+1 = (x^4-\sqrt{2}x^2+1)(x^4+\sqrt{2}x^2+1)$? $\endgroup$
    – sourisse
    Nov 1, 2012 at 20:05
  • $\begingroup$ @sourisse Ahh, no it doesn't, I simply didn't spot that one! Thanks! :) $\endgroup$ Nov 1, 2012 at 20:06
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    $\begingroup$ $u = x^2$ looks like it simplifies things a bit. $\endgroup$
    – user14972
    Nov 1, 2012 at 20:30

4 Answers 4

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These integrals are often wrapped up nicely by substitutions of the form: $$u=x^a\pm\frac{1}{x^a}$$ where $a$ is chosen appropriately. A little bit of playing around leads to the following: $$\int\frac{x^{5}-x}{x^{8}+1}dx=\int\frac{x^{3}\left(x^{2}-\frac{1}{x^{2}}\right)dx}{x^{4}\left(x^{4}+\frac{1}{x^{4}}\right)}=\int\frac{\left(x^{2}-\frac{1}{x^{2}}\right)dx}{x\left[\left(x^{2}+\frac{1}{x^{2}}\right)^{2}-2\right]}$$ Now let $$u=x^{2}+\frac{1}{x^{2}}$$ $$du=2\left(x-\frac{1}{x^{3}}\right)dx=2\frac{1}{x}\left(x^{2}-\frac{1}{x^{2}}\right)dx$$ Hence $$2I=\int\frac{du}{u^{2}-2}=\frac{1}{2\sqrt{2}}\int\frac{du}{u-\sqrt{2}}-\int\frac{du}{u+\sqrt{2}}=\frac{1}{2\sqrt{2}}\ln\left|\frac{u-\sqrt{2}}{u+\sqrt{2}}\right|$$ $$I=\frac{1}{4\sqrt{2}}\ln\left|\frac{x^{4}-\sqrt{2}x^{2}+1}{x^{4}+\sqrt{2}x^{2}+1}\right|$$

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  • $\begingroup$ Thank you, this is a useful substitution technique that I've never come across before! $\endgroup$ Nov 1, 2012 at 21:38
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You may try using factorization $x^8+1=(x^4-\sqrt{2}{\ }x^2+1)(x^4+\sqrt{2}{\ }x^2+1)$

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Decomposition in partial fractions. Easy calculation. http://en.wikipedia.org/wiki/Partial_fraction

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In real quadratics, $ x^8 + 1$ is $$ (x^2 - \; x \sqrt{2 + \sqrt 2} \; + 1) (x^2 + x \sqrt{2 + \sqrt 2} + 1) (x^2 - x \sqrt{2 - \sqrt 2} + 1) (x^2 + x \sqrt{2 - \sqrt 2} + 1) $$

I got this by finding $\cos \frac{\pi}{8}$ and $\sin \frac{\pi}{8}.$ On the other hand, you can easily see the relationship with the answer of M. Strochyk.

What this means is that, at the cost of square roots all over creation, partial fractions can be carried out completely, most likely including $\arctan$ and $\log$ terms, what ever else usually comes up.

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