this post was submitted on 05 Jul 2021
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[–] [email protected] 0 points 4 years ago (1 children)

Apparently it’s doable according to Wolfram

I guess you’d factor into irreducible quadratics and do partial fractions, I’m pretty sure you can always integrate (Ax+B)/(Cx^2 + Dx + E)-looking stuff, though I forget how some of the cases work out

But yeah it looks pretty shitty; even some nested radicals as factors, I am curious how those show up. Not great

[–] [email protected] 0 points 4 years ago* (last edited 4 years ago) (1 children)

Wolfram solution for anyone wondering. Admittedly I'm not the best when it comes to solving complicated integrals in general and I'm a bit out of practice as well, so I can feel my brain melting just looking at it lol.

[–] [email protected] 0 points 4 years ago (1 children)

If you change it to (x^5 - 1) you definitely can't do it though (probably, partial fractions won't work at least).

[–] [email protected] 1 points 4 years ago

tl;dr these 1/(x^n + k), or -k, problems are all doable with partial fractions because we know the roots.

Longer version: partial fractions would work, just with absolutely horrific algebra. All polynomials factor into a product of linear and irreducible quadratics.

We know all the roots of x^5 - 1 (the 5th roots of unity), so we can factor it completely over the complex numbers. Pair up the factors we get from complex conjugates, and their product should be quadratics with real (albeit really funky, involving the golden ratio because of the 5th power) coefficients.

Here, the irreducible factors of x^5 - 1 are (x-1), (x^2 - 2cos(2pi/5) + 1), and (x^2 - 2cos(4pi/5) + 1)

I actually like how many different things the problem ties together, but the algebra from here on out looks like a total slog; I set up start of the partial fractions decomposition, but I am not looking forward to the multiplications or system of equations. I might try and power through just because I do want to see what the integration looks like, dammit