Simple integrals for gravitational potential

MikeGomez
Messages
343
Reaction score
16

Homework Statement


Homework Equations



I need help solving intergral…
[tex]\int \frac{dx}{(a+x)^2}[/tex]

The Attempt at a Solution


I found the integral for…
[tex]\int \frac{dx}{(a^2+x^2)}[/tex] = 1/a arctan x/a

But I don’t know how to apply that to the original integral which is a little different

[tex]\int \frac{dx}{(a+x)^2} = \int \frac{dx}{(a^2+x^2+2ax)}[/tex]

I also need to solve the following integral
[tex]\int \frac{dx}{(a+b-x)^2}[/tex]


It’s not homework. The reason is that I want to work through the numbers that Rybczyk gives as equation 1 for gravitational potential in his paper “Gravitational Effect on Light Propagation”

http://www.mrelativity.net/Gravitat...ravitational Effects on Light Propagation.htm

I have simplified his equation in my post somewhat, as I already know how to separate the two terms separated by the minus sign, and assuming the gravitational constant G and the mass of the bodies are constant, I know that they can come out in front of the integral sign.

Also, I changed the variable names to more familiar ones. I hope the variable name substitutions helps rather than hinders. If not I'll have to rewrite this whole post using Rybczyk's exact variables..

Thanks.
 
on Phys.org
MikeGomez said:

Homework Statement


Homework Equations



I need help solving intergral…
[tex]\int \frac{dx}{(a+x)^2}[/tex]

The Attempt at a Solution


I found the integral for…
[tex]\int \frac{dx}{(a^2+x^2)}[/tex] = 1/a arctan x/a

But I don’t know how to apply that to the original integral which is a little different

[tex]\int \frac{dx}{(a+x)^2} = \int \frac{dx}{(a^2+x^2+2ax)}[/tex]

I also need to solve the following integral
[tex]\int \frac{dx}{(a+b-x)^2}[/tex]


It’s not homework. The reason is that I want to work through the numbers that Rybczyk gives as equation 1 for gravitational potential in his paper “Gravitational Effect on Light Propagation”

http://www.mrelativity.net/Gravitat...ravitational Effects on Light Propagation.htm

I have simplified his equation in my post somewhat, as I already know how to separate the two terms separated by the minus sign, and assuming the gravitational constant G and the mass of the bodies are constant, I know that they can come out in front of the integral sign.

Also, I changed the variable names to more familiar ones. I hope the variable name substitutions helps rather than hinders. If not I'll have to rewrite this whole post using Rybczyk's exact variables..

Thanks.
For the initial integral, use substitution, letting u = x+a .
 
Thanks Sammy. It looks to me like du = dx in this case. Is that right?

[tex]\int \frac{dx}{(a+x)^2} =[/tex]
[tex]\int {(a+x)^{-2}}{dx}[/tex]

u = a + x
du = dx

[tex]\int {(u)^{-2}}{du}=[/tex]
[tex]-\frac{1}{u} + c[/tex]
[edit] made correction (forgot the + c)

Substituting a + x for u gives

[tex]-\frac{1}{a+x} + c[/tex]

Is that right?
 
Last edited:
MikeGomez said:
Thanks Sammy. It looks to me like du = dx in this case. Is that right?

[tex]\int \frac{dx}{(a+x)^2} =[/tex]
[tex]\int {(a+x)^{-2}}{dx}[/tex]

u = a + x
du = dx

[tex]\int {(u)^{-2}}{du}=[/tex]
[tex]-\frac{1}{u}[/tex]
Substituting a + x for u gives
[tex]-\frac{1}{a+x}[/tex]

Is that right?

Yes, it's right.
 
Great, thanks Dick (quick reply!).

Will try same for the second part of original question.

NOTE: I editied post #3 to make a correction for the missing ( + c) for a constant.
 

Similar threads

  • · Replies 105 ·
4
Replies
105
Views
14K
Replies
3
Views
2K
  • · Replies 22 ·
Replies
22
Views
4K
  • · Replies 12 ·
Replies
12
Views
2K
Replies
7
Views
3K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 54 ·
2
Replies
54
Views
20K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K