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

  • Thread starter thebuttonfreak
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    Integral
In summary, the person is trying to solve an equation in which z-rx is the unknown. They have made a substitution for x and are unable to get an answer. They have tried several methods, including using Legendre polynomials, but are still having difficulty.
  • #1
thebuttonfreak
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int((z-r*x)/[z^2+r^2-2*z*r*x]^(3/2), x)
 
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  • #2
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:
  • #3
Would that be:
[tex]\int\frac{z-rx}{\sqrt[\frac{3}{2}]{z^{2}+r^{2}-2zrx}}dx[/tex]
 
  • #4
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.
 
  • #5
ps- how are you posting this mathematical notation? I am young but eager so please be patient with me guys.
 
  • #6
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.
 

1. What is an integral?

An integral is a mathematical concept that represents the area under a curve on a graph. It is used to find the total value of a function over a given interval. It is often used in calculus and other branches of mathematics to solve problems related to rates of change and accumulation.

2. Why do we need help with integrals?

Integrals can be complex and require a deep understanding of mathematical concepts. They also involve a lot of steps and can be time-consuming to solve. Seeking help with integrals can assist in solving problems efficiently and accurately.

3. How do you solve an integral?

To solve an integral, you need to find the anti-derivative of the function being integrated. This can be done by using integration techniques such as substitution, integration by parts, or trigonometric substitution. Once the anti-derivative is found, the integral can be evaluated at the given limits to find the final answer.

4. Can you give an example of solving an integral?

Sure, let's say we have the integral of 2x dx. To solve this, we first find the anti-derivative of 2x, which is x^2. Then, we plug in the limits of integration, let's say from 0 to 3. So the final answer would be (3)^2 - (0)^2 = 9.

5. Where can I find resources for help with integrals?

There are many online resources available for help with integrals, such as math forums, tutorial websites, and YouTube videos. Additionally, your local library or university may have textbooks or tutors available for assistance. It's also helpful to practice solving integrals on your own to improve your skills.

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