Simplifying (x^3)/((x^3)-2) to Integral Calculations and Complex Roots

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Homework Help Overview

The discussion revolves around the integral of the expression (x^3)/((x^3)-2), focusing on the simplification and subsequent steps needed for integration. Participants explore the complexities arising from the roots of the denominator and the implications of different interpretations of the expression.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss polynomial division and the subsequent expression, questioning how to proceed with integration after obtaining 1 + (2/((x^3)-2)). There is mention of the complexity of roots and the potential for confusion regarding the term 2 versus 2^(1/3). Some suggest using partial fractions as a method to simplify the integral.

Discussion Status

The discussion is ongoing, with participants sharing their attempts and confirming the challenges involved in simplifying the expression. There is acknowledgment of the difficulty in finding a straightforward solution, and some guidance has been offered regarding the use of partial fractions.

Contextual Notes

Participants note the ambiguity in the original expression and the importance of clarity in mathematical notation. The discussion also highlights the complexity of the roots involved in the denominator, which affects the approach to solving the integral.

transgalactic
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the the integral of

(x^3)/((x^3)-2)

i devided the polinomials and got

1+ (2/((x^3)-2))

and afterwards i tried to split the deminator
but i g0t really complicated roots
which are not alowing me to go further
how do i continue after i devided the polinomials
 
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nooo

its 2 in the power of(1/3) not 2

thats the big problem
how do i dolve that
 
If you want help, type it out w/o ambiguity or use LaTeX to avoid any confusion.
 
if we solve ny the formula then we need to substitue eache 2
with 2^(1/3)
 
ye Difference of cubes
but you wrote as if it was

x^3 -8


our question is:

1+ (2/((x^3)-2))
 
Last edited:
Well after looking at it and confirming with Mathematica, your numbers will be ugly. So you are probably on the right track, and I can't think of a slick way ... I tried several methods but didn't work out nicely at all.
 
Lol, I just did it ... it is really ugly in the sense of too many fractions ... I don't even want to type it up. It's pretty ez though, just use Partial Fractions and you're done.
 
thanks
 

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