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Homework Help: V^2 / v^2 + 4

  1. Feb 2, 2009 #1
    1. The problem statement, all variables and given/known data

    how do i integrate

    v^2 / v^2 + 4

    2. Relevant equations

    i understand this has something to do with arctan

    but if i use u substitution to let v=(u/2) so (on the bottom) it becomes (1/4)(1+(v/2)^2)

    there's still a v^2 on the top which the u substitution does not get rid of. Help :[

    3. The attempt at a solution
  2. jcsd
  3. Feb 2, 2009 #2
    this is a bit of a gues but why dont you write it as

    [itex]\int \frac{v}{2} \frac{2v}{v^2+4} dv[/itex]

    then do it by parts but for the 2nd term you can substitute [itex]u=v^2+4[/itex] giving [itex]\frac{du}{u}[/itex].

    you'll get logs not arctans which i reckon is true because the your integral isn't of the form [itex]\int \frac{dx}{x^2+a^2}[/itex]
  4. Feb 2, 2009 #3
    Use polynomial division to rewrite [tex] \frac{v^2}{v^2 + 4} [/tex] as [tex] 1 - \frac{4}{v^2 + 4}[/tex]. Can you integrate it now?
  5. Feb 2, 2009 #4
    ...or that lol!

    would't my method work as well?
  6. Feb 2, 2009 #5
    Doing it your way, you'd have to evaluate [tex] \int ln(x^2+4) [/tex] which evaluates to

    [tex] 4 tan^{-1}\left(\frac{x}{2}\right) + x \left(ln(x^2+4) - 2\right) [/tex]

    My way seems to be easier.

    Let me summarize what happens your way, I switched v's to x's.

    [tex] \int \frac{x^2}{x^2+4} dx = \int \frac{x}{2} \frac{2x}{x^2+4} dx = \frac{x}{2} ln(x^2+4) - \frac{1}{2} \int ln(x^2+4) dx = \frac{x}{2} ln(x^2+4) - \frac{1}{2} \left(4 tan^{-1}\left(\frac{x}{2}\right) + x \left(ln(x^2+4) - 2\right)\right) [/tex]

    Now obviously [tex] \frac{x}{2} ln(x^2+4) [/tex] will cancel and you will be left with

    [tex] x - 2 tan^{-1}\left(\frac{x}{2}\right) [/tex]

    This is the exact same answer you get if you do it my way:

    [tex] \int \frac{x^2}{x^2+4} dx = \int 1 - \frac{4}{x^2+4} dx = x - 4\left(\frac{1}{2} tan^{-1}\left(\frac{x}{2}\right)\right) = x - 2 tan^{-1}\left(\frac{x}{2}\right)[/tex]
    Last edited: Feb 2, 2009
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