Integration Tricks for Homework Statement on Electrodynamics Problem

[bjnartowt]

Homework Statement

Hi all, as part of an electrodynamics-problem, I encountered:

$\int {\frac{x}{{{{({x^2} + 1)}^3}}} \cdot dx}$

I know how to get the answer with my computer, but I wanted to know how to do it by hand. It’s getting to the point here in physics-grad-school that I need to know how to do this stuff “off the cuff”, so to speak.

Homework Equations

The Attempt at a Solution

Well...this is part of a solution of a bigger problem. I know there's some sort of trig substitution or trig identity. It happens that x = r/d, where "r" is an in-plane polar distance (x^2 + y^2), and "d" is distance above the plane, and I rearranged the integral I got into getting the dimensionless length scale x = r/d ... and by the above geometry, if you consider theta to be the angle between "r" and "d", then r/d = tan(theta). But:

dx = d(r/d) = d(tan(theta)) = (1 + tan(theta)^2)*d(theta)

That's just a mess.

 

[kbaumen]

Am I missing something? Or does it really just work with u=x^2+1?

 

[╔(σ_σ)╝]

The above coment is correct. This is a simple substitution of $$u=x^{2}+1$$. Then the next step is a simple interation of a polynomial of degree -3.

 

[kbaumen]

The above coment is correct. This is a simple substitution of $$u=x^{2}+1$$. Then the next step is a simple interation of a polynomial of degree -3.

I assume the dot there is an accident then, it doesn't mean anything.

 

[╔(σ_σ)╝]

Well it is obvious that the function does not define a vector field. I believe it means nothing.

 

[bjnartowt]

Wow...u = x^2 + 1 makes it easy-peezee, lemon squeezy. ::blush:: guess I missed it. I learned how to integrate stuff 7 years ago, and I've made the mistake of letting a computer do it for me most of the time. Bah.

 

[gomunkul51]

u = 1+x^2
du = 2x dx

-> integral = 0.5*(1/u^3)du

 

[Mark44]

The above coment is correct. This is a simple substitution of $$u=x^{2}+1$$. Then the next step is a simple interation of a polynomial of degree -3.

I assume the dot there is an accident then, it doesn't mean anything.

It does mean something - it's the period at the end of the sentence.

 

[╔(σ_σ)╝]

44 I think he was referring to the dot in OP's post, the one between his function and dx :-).

 

[bjnartowt]

I use dots to make scalar-multiplication clear. I recognize it as "it's ok to assume this kind of multiplication commutes", rather than the multiplication of things like operators and cross products.

 

[╔(σ_σ)╝]

Why,by convention dot is used for other things other than regular scalar multiplication. It might prove to be confusing for some people. The truth is that I originally didn't notice it.

 

[Mark44]

The above coment is correct. This is a simple substitution of $$u=x^{2}+1$$. Then the next step is a simple interation of a polynomial of degree -3.

I assume the dot there is an accident then, it doesn't mean anything.

44 I think he was referring to the dot in OP's post, the one between his function and dx :-).

I don't think so. kbaumen quoted your post (the first one above), not the OP with the integral in it.

 

[Mark44]

Well it is obvious that the function does not define a vector field. I believe it means nothing.

In this case, I believe kbaumen was referring to the period at the end of the sentence. Dots are often used for multiplication, especially with constants, as in
$$3 \cdot 5 = 15$$.

The first dot indicates multiplcation; the last one is the period at the end of the sentence.

 

[kbaumen]

Sorry for not making it clear, but I was indeed referring to the dot between the function and dx.