Need help with this integral!

1. Jan 23, 2007

thebuttonfreak

int((z-r*x)/[z^2+r^2-2*z*r*x]^(3/2), x)

2. Jan 23, 2007

cristo

Staff Emeritus
Is this the integral you want to solve? $$\int\frac{z-rx}{(z^2+r^2-2zrx)^{3/2}}dx$$

Last edited: Jan 23, 2007
3. Jan 23, 2007

arildno

Would that be:
$$\int\frac{z-rx}{\sqrt[\frac{3}{2}]{z^{2}+r^{2}-2zrx}}dx$$

4. Jan 23, 2007

thebuttonfreak

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.

5. Jan 23, 2007

thebuttonfreak

ps- how are you posting this mathematical notation? I am young but eager so please be patient with me guys.

6. Jan 23, 2007

cristo

Staff Emeritus
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.