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Integrate x^3sqrt(x^2+4)

  1. Jan 16, 2010 #1
    1. The problem statement, all variables and given/known data

    Integrate x^3sqrt(x^2+4)




    The attempt at a solution

    I have no idea how to substitute this integral in my favor. Can someone please set me on the right track? Thanks
     
  2. jcsd
  3. Jan 16, 2010 #2

    Dick

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    Substitute u^2=x^2+4. It's actually not so bad.
     
  4. Jan 17, 2010 #3
    Thanks for the help Dick.

    I did what you suggested but I didn't quite get the correct answer.

    Here are my steps:

    [tex]\int[/tex]x^3[tex]\sqrt{(x^2+4)}[/tex]dx = [tex]\int[/tex](x^3(x^2+4))dx/[tex]\sqrt{(x^2+4)}[/tex]

    u^2 = x^2+4 ; du = xdx/[tex]\sqrt{(x^2+4)}[/tex]

    [tex]\int[/tex]x^2u^2du

    x^2 = u^2 - 4

    [tex]\int[/tex] (u^2-4)u^2du
    = 1/5 (x^2+4)^5/2 - 4/3(x^2+4)^3/2 + c
     
  5. Jan 17, 2010 #4

    Mark44

    Staff: Mentor

    Starting from [itex]\int x^3 (x^2 + 4)^{1/2}dx[/itex]

    make all your substitutions, using
    u^2 = x^2 + 4 ==> x^2 = u^2 - 4 ==> x = (u^2 - 4)^(1/2)
    and udu = xdx

    Your integral [itex]\int (u^2 - 4) u^2 du[/itex] shows that you have not substituted correctly. You should get two terms in u, both with fractional exponents.
     
  6. Jan 17, 2010 #5
    I'm sorry but I'm really clueless.
    What step in my substitution did I do wrong?
     
  7. Jan 17, 2010 #6

    Dick

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    You didn't do anything wrong. Your answer is right. But there are other ways to write the answer. What answer are you expecting?
     
  8. Jan 17, 2010 #7

    Mark44

    Staff: Mentor

    Your work is correct - it's mine that is in error. Sorry for giving bad advice.
     
  9. Jan 17, 2010 #8
    Thank you both very much. The answer they give is [1/15(x^2+4)^3/2](3x^2-8) + c
     
  10. Jan 17, 2010 #9

    Dick

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    Just factor (x^2+4)^(3/2) out of both terms in your solution to get a form like that.
     
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