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

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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When you said 'nice' partial fraction decomposition, does that disbar $x^8+1 = (x^4-\sqrt{2}x^2+1)(x^4+\sqrt{2}x^2+1)$? – sourisse Nov 1 '12 at 20:05
@sourisse Ahh, no it doesn't, I simply didn't spot that one! Thanks! :) – Shaktal Nov 1 '12 at 20:06
$u = x^2$ looks like it simplifies things a bit. – Hurkyl Nov 1 '12 at 20:30
up vote 11 down vote accepted

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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Thank you, this is a useful substitution technique that I've never come across before! – Shaktal Nov 1 '12 at 21:38

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.

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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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