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Integrate ln(4+y^2)dy?

  1. Jan 16, 2007 #1
    Stuck at the end of a double integral, still have to integrate ln(4+y^2)dy

    Assuming I did the right first step. Original double integral is

    x/(x^2+y^2)

    Thanks!
     
  2. jcsd
  3. Jan 16, 2007 #2

    benorin

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    What are the bounds on x and y?
     
  4. Jan 16, 2007 #3
    the region R = [1,2] * [0,1]
     
  5. Jan 16, 2007 #4

    Hurkyl

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    Have you had any thoughts on integrating that? I see two obvious things to try:

    (1) Do what you normally do with integrals of logarithms.
    (2) Make a substitution.
     
  6. Jan 16, 2007 #5

    mjsd

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    eg. Integration by parts then trig sub.
     
  7. Jan 17, 2007 #6

    dextercioby

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    There's no need for substitution for that integral. Part integration once then a smart move in the numerator of the remaining integral and it's done.

    Daniel.
     
  8. Jan 17, 2007 #7
    is the integral we are talking about [tex]\int_{0}^{1}\int_{1}^{2}\frac{x}{x^2+y^2}dxdy[/tex]?

    i get [tex]\frac{1}{2}\int_{0}^{1}\left[\ln(4+y^2)-\ln(1+y^2)\right]dy[/tex]
    what should i do next? (edit: i got it. integration by parts.)
     
    Last edited: Jan 17, 2007
  9. Jan 17, 2007 #8

    Gib Z

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    Murshid, split the integral up, do them separately perhaps? Eg, say for [tex]\int ln(1+y^2) dy[/tex] we could let y^2 equal u. Find du/dy, easy. Then, solve for dy. Now substitute that value in. We end up with [tex]\int \frac{ln (u+1)}{2(u)^{1/2}} du [/tex]. Then some nice integration by parts and we are done?

    Takes a while though, I hope your patient.
     
  10. Jan 18, 2007 #9
    well i got it already. thanks anyway.
    but we can directly use integration by parts on this [tex]\int ln(1+y^2) dy[/tex] by letting [tex]u = \ln(1+y^2)[/tex] and [tex]dv = dy[/tex]
     
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