# Simple integrals for gravitational potential

1. Jan 7, 2013

### MikeGomez

1. The problem statement, all variables and given/known data
2. Relevant equations

I need help solving intergral…
$$\int \frac{dx}{(a+x)^2}$$

3. The attempt at a solution
I found the integral for…
$$\int \frac{dx}{(a^2+x^2)}$$ = 1/a arctan x/a

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

$$\int \frac{dx}{(a+x)^2} = \int \frac{dx}{(a^2+x^2+2ax)}$$

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

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.

2. Jan 7, 2013

### SammyS

Staff Emeritus
For the initial integral, use substitution, letting u = x+a .

3. Jan 7, 2013

### MikeGomez

Thanks Sammy. It looks to me like du = dx in this case. Is that right?

$$\int \frac{dx}{(a+x)^2} =$$
$$\int {(a+x)^{-2}}{dx}$$

u = a + x
du = dx

$$\int {(u)^{-2}}{du}=$$
$$-\frac{1}{u} + c$$
 made correction (forgot the + c)

Substituting a + x for u gives

$$-\frac{1}{a+x} + c$$

Is that right?

Last edited: Jan 7, 2013
4. Jan 7, 2013

### Dick

Yes, it's right.

5. Jan 7, 2013

### MikeGomez

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.