$\int\frac{1}{x^6+1}dx$ Integrating a rational function, help with factoring the denominator

Question

$$\int\frac{1}{x^6+1}\,dx$$

How to do this kind of integral, when one has a high degree polynomial in the denominator?

There is an idea in the book to add and subtract $x^2$ in the numerator, so one gets two integrals, but they are easy to do.

$$\int\frac{1}{x^6+1}\,dx = \int\frac{x^2+1}{x^6+1} \,dx - \int\frac{x^2}{x^6+1}\,dx$$

But how would I know that $x^6+1$ is divisible by $x^2+1$? Is there a method that doesn't rely on that division quirk?

Because $x^6+1=(x^2)^3+1^3$.... Because you have sum of cubes... $a^3+b^3=(a+b)(a^2-ab+b^2)$ So $x^6+1=(x^2+1)((x^2)^2-x^2+1)$ — randomgirl, Aug 05 '17 at 12:00
Another (perhaps more advanced way) would be to factor $x^6+1$ over the complex numbers, integrate the partial fractions with complex coefficients. — GEdgar, Aug 05 '17 at 13:13

score 2 · Accepted Answer · answered Aug 05 '17 at 12:01

2

Since $x^3+1=(x+1)(x^2-x+1)$, $x^6+1=(x^2+1)(x^4-x^2+1)$. Therefore, your final sum is equal to$$\int\frac1{x^4-x^2+1}\,\mathrm dx-\int\frac{x^2}{x^6+1}\,\mathrm dx$$

answered Aug 05 '17 at 12:01

José Carlos Santos

427,504

You Sir are such a genius. – Shocky2 Aug 05 '17 at 12:03
And what about 1/(x^4+1)? – Shocky2 Aug 05 '17 at 12:05
@Shocky2 What about it? I made no mention to it and neither did you. – José Carlos Santos Aug 05 '17 at 12:09
I ask if you could answer it too here, so I don't ask another question. – Shocky2 Aug 05 '17 at 13:25
2

@Shocky2 Multiple questions at once are usually frowned upon, and answerers hold no obligation to answer new questions on old questions. – Simply Beautiful Art Aug 05 '17 at 13:31
Excuse my unfamiliarness with the etiquette. Thank you for your time. – Shocky2 Aug 05 '17 at 13:32

score 1 · Answer 2 · answered Aug 05 '17 at 12:19

1

Next observe this. $$x^4 - x^2 + 1 = x^4 + 2x^2 + 1 - 3x^2 = (x^2 + 1)^2 - \left(x\sqrt{3}\right)^2.$$ Factor this difference of squares to factor the quartic into two irreducible quadratics. Then use a partial fractions expansion.

answered Aug 05 '17 at 12:19

ncmathsadist

49,383

score 1 · Answer 3 · answered Aug 05 '17 at 20:43

1

$\dfrac{1}{x^6+1}=\dfrac{1}{(x-1) (x+1) \left(x^2-x+1\right) \left(x^2+x+1\right)}=$

$= \dfrac{1}{6} \left(\dfrac{x}{x^2-x+1}-\dfrac{x}{x^2+x+1}-\dfrac{2}{x^2-x+1}-\dfrac{2}{x^2+x+1}+\dfrac{1}{x-1}-\dfrac{1}{x+1}\right) $

Hope this helps

answered Aug 05 '17 at 20:43

Raffaele

26,371

Dr. Sonnhard Graubner · Answer 4 · 2017-08-05T12:30:14.980

0

note that $$\frac{1}{1+x^6}=1/3\, \left( {x}^{2}+1 \right) ^{-1}+1/3\,{\frac {-{x}^{2}+2}{{x}^{4}- {x}^{2}+1}} $$

edited Aug 05 '17 at 12:30

answered Aug 05 '17 at 12:02

Dr. Sonnhard Graubner

95,283

$\int\frac{1}{x^6+1}dx$ Integrating a rational function, help with factoring the denominator

4 Answers4