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Integral with substitution

  1. Aug 9, 2011 #1
    1. The problem statement, all variables and given/known data
    find the integral from 1/3 to 1/2 of 1/x(x+1)(x-1)(x^2+1)


    2. Relevant equations



    3. The attempt at a solution
    expanded and simplified:
    1/x^5-x

    it gives a hint suggesting i use a substitution here, but i can't find an appropriate one.. Help?
     
  2. jcsd
  3. Aug 9, 2011 #2

    I like Serena

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    Welcome to PF, Shannabel! :smile:

    Perhaps try for instance [itex]u=x^4-1[/itex]?
     
  4. Aug 9, 2011 #3
    Are there suppose to be terms in the numerator besides 1? From what you wrote I would assume no but I wanted to make sure...
     
  5. Aug 9, 2011 #4
    nope, just a 1
     
  6. Aug 9, 2011 #5

    I like Serena

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    Let me clean that up:
    [tex]\int_{1/3}^{1/2} {1 \over x(x+1)(x-1)(x^2+1)} dx[/tex]

    Click the Quote button to see what I did (it is called LaTeX). :wink:
     
  7. Aug 9, 2011 #6
    i don't see how that will work.. since then du = 4x^3dx
     
  8. Aug 9, 2011 #7
    Ya then I would stick with I like Serena's suggestion if you haven't learned partial fraction decomposition yet. Expanding is kind of tedious and there are some very common integrals than can be evaluated once you resolve that into partial fractions.
     
  9. Aug 9, 2011 #8

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    Yep. Keep going! :smile:
     
  10. Aug 9, 2011 #9
    then i get
    du/4x^4(u) ... but i can't integrate that?
     
  11. Aug 9, 2011 #10

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    Well, you still have x in your expression.
    x should be eliminated... what would you get then?
     
  12. Aug 9, 2011 #11
    how do i eliminate the x?
     
  13. Aug 9, 2011 #12

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    Can you solve x from u=x4-1?
    And substitute that?
     
  14. Aug 9, 2011 #13
    do you mean like x^4 = u+1?
     
  15. Aug 9, 2011 #14

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    Yep! :smile:
     
  16. Aug 9, 2011 #15
    then i get
    du/u^2+u from 1/3 to 1/2
    which is lnu^2 + ln u from 1/3 to 1/2
    =2lnu+lnu
    =3lnu from 1/3 to 1/2

    u = x^4-1
    so when x = (1/2), u = -15/16
    and when x = (1/3), u = -80/81

    so from above i have
    3ln(-15/16)-3ln(-80/81)

    where did i go wrong? lol
     
  17. Aug 9, 2011 #16
    then i get
    du/u^2+u from 1/3 to 1/2
    which is lnu^2 + ln u from 1/3 to 1/2
    =2lnu+lnu
    =3lnu from 1/3 to 1/2

    u = x^4-1
    so when x = (1/2), u = -15/16
    and when x = (1/3), u = -80/81

    so from above i have
    3ln(-15/16)-3ln(-80/81)

    where did i go wrong? lol
     
  18. Aug 9, 2011 #17

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    Not so fast! :eek:

    Let's start with:

    And I'd really like some parentheses here, otherwise it becomes very confusing. :confused:

    So you had: du/(u^2+u)

    and now you're saying this integrates to: ln(u^2) + ln u

    Let's check that...
    Could you perhaps calculate the derivative of that last expression?
     
  19. Aug 9, 2011 #18
    ohhh
    i should have had
    (du/u^2)+(du/u)
    which integrates to
    -1/u + ln u?
     
  20. Aug 9, 2011 #19

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    Hmm, let's take a step back. :uhh:

    You had:
    So what do you get after you substitute (with the proper parentheses please)?
     
  21. Aug 9, 2011 #20
    du/[4(u+1)(u)]?
     
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