Could you help me to integrate

$$ \int{\frac{dx}{x^3+x^8}} $$

I've tried partial fraction decomposition but got $(x^4-x^3+x^2-x+1)$ as the last term when factored the denominator.

Thank you.

  • $\begingroup$ $x^4-x^3+x^2-x+1$ has complex roots equal to the primitive $10$th roots of unity. You might be able to do partial fractions with that. It will be ugly, however. $\endgroup$ – Thomas Andrews Mar 20 '15 at 23:43
  • $\begingroup$ I mean, you could break out the quartic formula to factor it, but yeah....ouch. Why do you want this one? $\endgroup$ – Alan Mar 20 '15 at 23:57
  • $\begingroup$ I thought I can make the last term the square of quadratic polynomial, but I can't. $\endgroup$ – Seva Mar 20 '15 at 23:57
  • $\begingroup$ You can factor $x^4-x^3=x^2-x+1=(x^2-2\cos(\pi/5)x +1)(x^2-2\cos(3\pi/5)x+1)$. That lets you get the two $\tan^{-1}$ terms in the Wolfram Alpha answer that C.T. got. $\endgroup$ – Thomas Andrews Mar 20 '15 at 23:59
  • $\begingroup$ Thank you very much! I'll try! $\endgroup$ – Seva Mar 21 '15 at 0:02

The best way to get the answer given by Wolfram Alpha is to write:

$$\frac{1}{x^3+x^8} = \frac{1}{x^3} -\frac{x^2}{x^5+1}$$

Then solve:

$$\frac{x^2}{1+x^5}=\frac{a}{x+1} + \frac{bx+c}{x^2-2\cos(\pi/5)x+1}+\frac{dx+e}{x^2-2\cos(3\pi/5)x+1}$$

Since $\sin\pi/5$ and $\cos\pi/5$ are in terms of $\sqrt{5}$, you can rewrite the answer in terms of $\sin(\pi/5),\cos(\pi/5),\sin(3\pi/5),\cos(3\pi/5)$. It still won't be prtty.

  • $\begingroup$ Wow, what is this factoring trick? I've never seen such a thing before. $\endgroup$ – mathamphetamines Mar 21 '15 at 0:38
  • $\begingroup$ Which factoring trick? The roots of $x^5+1$ are $-1$ and the primitive $10$th roots of unity, which can be written as $\cos(\pi/5)\pm i\sin(\pi/5)$ and $\cos(3\pi/5)\pm i(\sin(3\pi/5)$. $\endgroup$ – Thomas Andrews Mar 21 '15 at 0:53
  • $\begingroup$ I see. Thanks for the explanation! $\endgroup$ – mathamphetamines Mar 21 '15 at 1:02

You can see the answer computed by Wolfram alpha below. I wouldn't try to do this one by hand though.

enter image description here

  • $\begingroup$ I've tried already and the answer frightened me. $\endgroup$ – Seva Mar 20 '15 at 23:58

$x^3+x^8=x^3(1+x^5)$ and $x^5=-1=e^{in\pi}$, ($n$ odd), when $x=e^{\frac{in\pi}{5}}$ for $n=1,3,5,7$, and $9$.

Note that there are two real roots, $x=0$ (a thrice repeated root) and $x=-1$, along with two sets of complex conjugate roots.

Equipped with the roots, partial fraction expansion is a straightforward, brute force approach.

  • $\begingroup$ Sorry, but why $x^5=e^{in\pi}$, but not $e^{i\pi}$? $\endgroup$ – Seva Mar 21 '15 at 0:50
  • $\begingroup$ That will give you only 1 root, not all of them. And the expression for odd $n$ is valid. $\endgroup$ – Mark Viola Mar 21 '15 at 4:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.