Efficient Integral Solver for (z-r*x)/[z^2+r^2-2*z*r*x]^(3/2), x)

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Discussion Overview

The discussion revolves around solving the integral \(\int \frac{z - r x}{(z^2 + r^2 - 2 z r x)^{3/2}} \, dx\). Participants explore various methods and substitutions to tackle this integral, while also addressing the use of mathematical notation in the forum.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant presents the integral to be solved and requests assistance.
  • Another participant asks for the original poster's work to adhere to forum rules before providing help.
  • A participant suggests a substitution \(u = z^2 + r^2 - 2zrx\) but expresses difficulty in canceling terms and solving the integral.
  • There is mention of using Legendre polynomials as a potential method for solving the integral, although the original poster prefers not to use that technique.
  • One participant offers a suggestion to replace \(x\) using the substitution, indicating a possible direction for further exploration.
  • Another participant claims to have obtained an answer using an integrator tool, suggesting that the original poster may have made an error in inputting the integral.
  • A participant provides guidance on how to post mathematical notation correctly in the forum.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the solution method or the correctness of the approaches discussed. Multiple competing views and methods remain present in the conversation.

Contextual Notes

Participants express uncertainty regarding the effectiveness of their substitutions and the correctness of their mathematical input into integrator tools. There is also a lack of clarity on the best approach to solve the integral without relying on specific techniques like Legendre polynomials.

thebuttonfreak
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int((z-r*x)/[z^2+r^2-2*z*r*x]^(3/2), x)
 
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Is this the integral you want to solve? [tex]\int\frac{z-rx}{(z^2+r^2-2zrx)^{3/2}}dx[/tex]

Could you please show your efforts first, since PF rules state that we must see your work before we can give help you.
 
Last edited:
Would that be:
[tex]\int\frac{z-rx}{\sqrt[\frac{3}{2}]{z^{2}+r^{2}-2zrx}}dx[/tex]
 
sure, i made the substitution u=z^2+r^2-2zrx, but got since du/dx=2zr and i was unable to cancel the x on top. also i let u =(z^2+r^2-2zrx)^3/2, i was unable to solve it. I tried maple, mathworld integrator but was unable to get an answer. I did that so i could at least see what direction to go in. Don't want the solution straight out but help on what direction to go would be great. I know it can be solved using Legendre polynomials but I want to solve it without use of that technique.




cristo said:
Is this the integral you want to solve? [tex]\int\frac{z-rx}{(z^2+r^2-2zrx)^{3/2}}dx[/tex]

Could you please show your efforts first, since PF rules state that we must see your work before we can give help you.
 
ps- how are you posting this mathematical notation? I am young but eager so please be patient with me guys.
 
thebuttonfreak said:
sure, i made the substitution u=z^2+r^2-2zrx, but got since du/dx=2zr and i was unable to cancel the x on top. also i let u =(z^2+r^2-2zrx)^3/2, i was unable to solve it. I tried maple, mathworld integrator but was unable to get an answer. I did that so i could at least see what direction to go in. Don't want the solution straight out but help on what direction to go would be great. I know it can be solved using Legendre polynomials but I want to solve it without use of that technique.

If you use the first substitution then you can replace x by noting that x=(z^2+r^2-u)/2zr. I've not done it, but this may help.

I get an answer when I put it into mathworld's integrator, so you may have typed it in wrong.

To see the mathematical notation, just click on one of the equations to see the code. Then include it in normal text with [ tex ] before and [ /tex ] after the code (without the spaces inside the brackets). See here for the LaTex tutorial.
 

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