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

Answer 1

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$$\int\frac1{x^4-x^2+1}\,\mathrm dx-\int\frac{x^2}{x^6+1}\,\mathrm dx$$

Answer 2

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

Answer 3

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.

Answer 4

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

$= \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