kramberol

2022-07-02

Partial fraction integration with unclear roots
Let's look at a simple example like $\frac{1}{{x}^{3}+2x+1}$ here. We know that the denominator has a real root between $0$ and $-1$ (could go closer, but that's not the point). By the concept of slope of a cubic being always positive or always negative, now I need to turn it into ($ax+b/$quadratic$\right)+\left(1/$linear form$\right)$.. For that, I'm going to need that root. So, in general, like the case of cubic where the roots are hard to calculate (with non real ones being in pairs as per theory), is there another way to factorise the denominator into linear and quadratics, or do I need to find suitable substitutions to find the actual root of equation like in quadratic which is near about impossible during examination time? (I can take care of repeated roots and other fundamental things related to partial fractions.)

Jayvion Mclaughlin

It's not a trivial amount of work but at least this is a depressed cubic already, and you can use something like Vieta's substitution. Let $x=w-\frac{2}{3w}$ to get:
${x}^{3}+2x+1={w}^{3}+1-\frac{8}{27{w}^{3}}=0.$
Multiply by ${w}^{3}$:
${w}^{6}+{w}^{3}-\frac{8}{27}=0,$
with a larger root
${w}^{3}=\frac{-1+\sqrt{1+\frac{32}{27}}}{2}\equiv C.$
Then the three cube roots of ${w}^{3}$ are $\sqrt[3]{C},\left(±\frac{\sqrt{3}i}{2}-\frac{1}{2}\right)\sqrt[3]{C},$ from which you can get the corresponding $x$'s.

Do you have a similar question?