1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Integration Tricks

  1. Aug 26, 2010 #1
    1. The problem statement, all variables and given/known data
    Hi all, as part of an electrodynamics-problem, I encountered:

    [itex]\int {\frac{x}{{{{({x^2} + 1)}^3}}} \cdot dx} [/itex]

    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.

    2. Relevant equations

    3. 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. :-p
  2. jcsd
  3. Aug 26, 2010 #2
    Am I missing something? Or does it really just work with u=x^2+1?
  4. Aug 26, 2010 #3
    The above coment is correct. This is a simple substitution of [tex]u=x^{2}+1[/tex]. Then the next step is a simple interation of a polynomial of degree -3.
  5. Aug 26, 2010 #4
    I assume the dot there is an accident then, it doesn't mean anything.
  6. Aug 26, 2010 #5
    Well it is obvious that the function does not define a vector field. I believe it means nothing.
  7. Aug 26, 2010 #6
    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.
  8. Aug 26, 2010 #7
    u = 1+x^2
    du = 2x dx

    -> integral = 0.5*(1/u^3)du
  9. Aug 26, 2010 #8


    Staff: Mentor

    It does mean something - it's the period at the end of the sentence.
  10. Aug 26, 2010 #9
    44 I think he was referring to the dot in OP's post, the one between his function and dx :-).
  11. Aug 26, 2010 #10
    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.
  12. Aug 26, 2010 #11
    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.
  13. Aug 26, 2010 #12


    Staff: Mentor

    I don't think so. kbaumen quoted your post (the first one above), not the OP with the integral in it.
  14. Aug 26, 2010 #13


    Staff: Mentor

    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
    [tex]3 \cdot 5 = 15[/tex].

    The first dot indicates multiplcation; the last one is the period at the end of the sentence.
  15. Aug 29, 2010 #14
    Sorry for not making it clear, but I was indeed referring to the dot between the function and dx.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook