Integrate ln(4+y^2)dy?

  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


  2. jcsd
  3. benorin

    benorin 1,026
    Homework Helper

    What are the bounds on x and y?
  4. the region R = [1,2] * [0,1]
  5. Hurkyl

    Hurkyl 15,987
    Staff Emeritus
    Science Advisor
    Gold Member

    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. mjsd

    mjsd 858
    Homework Helper

    eg. Integration by parts then trig sub.
  7. dextercioby

    dextercioby 12,327
    Science Advisor
    Homework Helper

    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.

  8. 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. Gib Z

    Gib Z 3,347
    Homework Helper

    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. 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]
Know someone interested in this topic? Share this thead via email, Google+, Twitter, or Facebook

Have something to add?