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U substitution. Why -1/x^2 is my du?

  1. Jan 26, 2014 #1
    1. The problem statement, all variables and given/known data

    ∫(1/x^(2))(3+1/x)^(3)

    2. Relevant equations

    U substitution is the way to go here

    3. The attempt at a solution

    My problem is that I cant figure my du and what is next. I know which one it is but I don't know the reason for it.

    u=3+1/x
    du= I chose ln|x| first but that got me nowhere. My buddy told me (txt) me it is -1/x^(2) wich eventually takes you to -(u)^4/4 but i cant figure out what happened in between.

    thanks for the hand.
     
  2. jcsd
  3. Jan 26, 2014 #2
    BTW i just need someone to explain me how to choose du. I'm fine after that
     
  4. Jan 26, 2014 #3

    SammyS

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    (Use the superscript, X2, icon for exponents.)
    First of all, it's a big help to include the dx with your integral:

    (1/x2)(3+1/x)3 dx

    It can look even better in LaTeX .

    ## \displaystyle
    \int \frac{1}{x^2}\left(3+\frac{1}{x}\right)^3\, dx
    ##

    What is the derivative of ## \displaystyle \left(3+\frac{1}{x}\right)^3 \ ?##

    (That won't give the answer directly, but it may lead you to it.)
    .
     
  5. Jan 26, 2014 #4
    alright, with the chain rule I get 3(3+1/x)^2*-1/x^2

    -1/x^2 would be my du then. (I was deriving it wrong before) so now that I have this minus sign I put it outside the Integrate sign. after that I'm good. Thanks!

    So if instead of a -1 i would have a 3 do I deal with it as if it were k and put it outside of the integral?

    BTW I've just downloaded LaTex
     
  6. Jan 26, 2014 #5

    Dick

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    You can always move a constant outside of an integral. And you didn't need to download LaTex. It's built into this site. Try looking here https://www.physicsforums.com/showpost.php?p=3977517&postcount=3
     
  7. Jan 26, 2014 #6
    [itex]∫(1/x^2)(3+1/x)^3[\itex] test
     
  8. Jan 26, 2014 #7

    Dick

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    [itex]∫(1/x^2)(3+1/x)^3[/itex] The slash in front of \itex is in the wrong direction. It should be /itex.
     
  9. Jan 26, 2014 #8

    SammyS

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    You don't seem to like dx !!!

    du would be -1/x2dx , multiplied by some constant.

    Since you end up with (3+1/x)2 , and you need (3+1/x)^3, you must need to start with a different power on the (3 + 1/x) factor.
     
  10. Jan 26, 2014 #9
    true, I have to pay more attention with writing dx

    My final answer is:

    [itex]-((3+1/x)4/4) + c[/itex]

    thanks for the help!
     
  11. Jan 26, 2014 #10

    Mark44

    Staff: Mentor

    You can't mix LaTeX with the <sup> tags. Here's what you wrote, fixed.
    ##-((3 + 1/x)^4/4) + C##

    My LaTeX script looks like this:
    -((3 + 1/x)^4/4) + C
     
  12. Jan 26, 2014 #11

    Dick

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    Right, [itex]-((3+1/x)^4/4) + c[/itex]. If you are wondering why it didn't TeX, don't use the shortcut superscripts and stuff inside Tex. Use ^ to get superscripts.
     
    Last edited: Jan 26, 2014
  13. Jan 26, 2014 #12
    [itex]-((3 + 1/x)^4/4) + C[/itex] lets see now

    Thank y'all for the help
     
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