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Another double integral problem

  1. Jan 6, 2010 #1
    1. The problem statement, all variables and given/known data
    sketch the region of integration, and evaluate the integral by choosing the best order of integration

    2. Relevant equations
    integration by parts

    3. The attempt at a solution
    after sketching the graph and changing the limits ive got to:



    [tex]u = {y^3}[/tex]

    [tex]u' = 3{y^2}[/tex]

    [tex]v = ln({y^4}+1)[/tex]

    [tex]v' = \frac{1}{{y^4}+1}[/tex]


    [tex]u = ln({y^4}+1)[/tex]

    [tex]u' = \frac{1}{{y^4}+1}[/tex]

    [tex]v = {y^3}[/tex]

    [tex]v' = 3{y^2}[/tex]

    [tex] = {y^3}ln({y^4}+1)-\int{y^3}\frac{1}{{y^4}+1}[/tex]

    And this is where i get stuck as its just an infinite loop of integration by parts, any ideas?
  2. jcsd
  3. Jan 6, 2010 #2


    User Avatar
    Homework Helper

    From here


    Just sub u=y4+1



    [tex]\frac{d}{dy}[ln(y^4+1)] \neq \frac{1}{y^4 +1}[/tex]
  4. Jan 7, 2010 #3
    Ok ive done that and i got the right answer :) but what is [tex]\int\frac{1}{{y^4}+1}[/tex] if its not [tex]ln({y^4}+1)[/tex] ?

    I remember how to integrate ln functions it using the joke:
    Q: whats the integral of 1 over cabin?
    A: log cabin
  5. Jan 7, 2010 #4
    That integral is too hard to be calculated by hand. But I've prepared it for you to not make mistakes like that: [tex]1/8\,\sqrt {2}\ln \left( {\frac {{x}^{2}+x\sqrt {2}+1}{{x}^{2}-x \sqrt {2}+1}} \right) +1/4\,\sqrt {2}\arctan \left( x\sqrt {2}+1\right) +1/4\,\sqrt {2}\arctan \left( x\sqrt {2}-1 \right) [/tex].

  6. Jan 7, 2010 #5
    I'll have to take your word on that one, lets just hope it doesn't come up in my exam!
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