How Do You Simplify Complex Root Integrals?

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Homework Help Overview

The discussion revolves around the simplification of complex root integrals, specifically focusing on the integral of a rational function involving square and cube roots.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore various substitution methods to simplify the integral, with some expressing frustration over the emergence of additional roots. Questions arise regarding the effectiveness of different substitutions and the handling of derivatives.

Discussion Status

Participants are actively engaging with different substitution strategies, with hints and suggestions being shared. There is an ongoing exploration of the implications of these substitutions, but no consensus has been reached on a definitive approach.

Contextual Notes

Some participants note constraints related to the powers involved in their substitutions, questioning the feasibility of certain transformations due to the complexity introduced by roots in the differential expressions.

transgalactic
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i tried this:
[tex] \int \frac{1-\sqrt{x+1}}{1+\sqrt[3]{x+1}}=\int \frac{1-\sqrt{x+1}}{1+\sqrt[3]{x+1}}*\frac{1+\sqrt{x+1}}{1+\sqrt{x+1}}*\frac{1-\sqrt[3]{x+1}+(x+1)^{\frac{3}{2}}}{1-\sqrt[3]{x+1}+(x+1)^{\frac{3}{2}}}[/tex]
but when i got read of 2 roots i got another two roots which are more complicated
??
 
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transgalactic said:
[tex] \int \frac{1-\sqrt{x+1}}{1+\sqrt[3]{x+1}}[/tex]

Hi transgalactic! :smile:

Hint: substitute! :wink:
 
i tried t=(x+1)^1/3
but it creates anoted roots in the dt
??
 
transgalactic said:
i tried t=(x+1)^1/3
but it creates anoted roots in the dt
??

uhh? :confused:

dx = … ?

anyway, (x+1)1/6 might be easier.
 
but i don't have members of 1/6 power
??
 
Sure you do. a^(1/2) = a^(1/6)^3, and b^(1/3) = b^(1/6)^2.
 

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