# Help integrate $\frac{1}{x^3+x^8}$

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.

• $x^4-x^3+x^2-x+1$ 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. – Thomas Andrews Mar 20 '15 at 23:43
• I mean, you could break out the quartic formula to factor it, but yeah....ouch. Why do you want this one? – Alan Mar 20 '15 at 23:57
• I thought I can make the last term the square of quadratic polynomial, but I can't. – Seva Mar 20 '15 at 23:57
• 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. – Thomas Andrews Mar 20 '15 at 23:59
• Thank you very much! I'll try! – 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.

• Wow, what is this factoring trick? I've never seen such a thing before. – mathamphetamines Mar 21 '15 at 0:38
• 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 $\cos(3\pi/5)\pm i(\sin(3\pi/5)$. – Thomas Andrews Mar 21 '15 at 0:53
• I see. Thanks for the explanation! – 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. • I've tried already and the answer frightened me. – 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.

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