How do you calculate the integral dz/(z^2+h^2)^(2/3)?

[NadAngel]

I am trying to solve a problem I saw in a book about electrostatics. In the book the solution to the integral ∫dz/(z^2+h^2)^(3/2) is z/(h^2*(z^2+h^2)^(1/2) but I can't solve the integral by myself. I have tried integration by substitution but I can't seem to solve it.

u=z^2+h^2
du=2zdz
dz=du/2z

This does not yield a correct result.

PS: h is a constant, so you only need to integrate it by dz.

 

[chiro]

Hey NadAngel and welcome to the forums.

That answer is not correct and you can differentiate the answer to see if it equals the expression inside the integral from the fundamental theorem of calculus. I would try this and do this first.

For the substitution though wikipedia has a good list of integral tables:

http://en.wikipedia.org/wiki/List_of_integrals_of_rational_functions

 

[JJacquelin]

Hi !
I doubt that you could integrate it with this method.
I think that this integral cannot be expressed as a combination of a finite number of usual fuctions. The closed form will probably requires an Hypergeometric function, or more simply, an Incomplete Beta function.

 

[NadAngel]

Thank you for the replies. It seems I have made a typo in the post and the correct integral of interest is:
∫dz/(z^2+h^2)^(3/2)

The solution in the book is the same: z/(h^2*(z^2+h^2)^(1/2)

Can this be soved?

 

[Curious3141]

Thank you for the replies. It seems I have made a typo in the post and the correct integral of interest is:
∫dz/(z^2+h^2)^(3/2)

The solution in the book is the same: z/(h^2*(z^2+h^2)^(1/2)

Can this be soved?

Yes, quite simply too.

First rearrange algebraically to get:

$$\frac{1}{h^3}\int {[1 + {(\frac{z}{h})}^2]}^{-\frac{3}{2}} dz$$

Then make the substitution $$\frac{z}{h} = \tan \theta$$.

Can you take it from there? Remember your trig identities.