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How to integrate this?

  1. Mar 5, 2014 #1
    View attachment 67272

    I need some help, thank you:-)
     
  2. jcsd
  3. Mar 5, 2014 #2
    £(x^(1/2))/(x-1)dx ,the upper limit is 4, the lower limit is 3
     
  4. Mar 5, 2014 #3

    SteamKing

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    What exactly does this mean? Why are you trying to integrate Pounds sterling?
     
  5. Mar 5, 2014 #4

    II'm sorry "£" here represent the integration symbol ....I can't type that on my phone...
     
  6. Mar 5, 2014 #5

    epenguin

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    You've got this awkward square root so a natural way to try and get rid of it would be the substitution y = x2 and this seems to bring it back to hopefully familiar or recognisable things.
     
  7. Mar 5, 2014 #6

    vanhees71

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    I guess the integral is
    [tex]I=\int \mathrm{d} x \frac{\sqrt{x}}{x-1}.[/tex]
    Then I'd substitute
    [tex]u=\sqrt{x}, \quad x=u^2 \; \mathrm{d} x = \mathrm{d} u 2 u.[/tex]
    Mod note: I removed part of this post as it was too much help.
     
    Last edited by a moderator: Mar 5, 2014
  8. Mar 5, 2014 #7

    arildno

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    Note that vanhees71's suggestion also suggests a more general thinking on square roots.
    Essentially, square roots (not to mention other types of roots!!) are nasty, and a good, general procedure is to seek to get rid of them by setting a pesky square root expression equal to a new variable, hoping that your problem disappears, say with productions of squares, rather than square roots.
    -----------------------------------------
    In other types of square root problems, the trick is to make a perfect square out of the expression beneath the square root sign, so that this new square precisely cancels out the bothersome root sign.
     
  9. Mar 5, 2014 #8

    Mark44

    Staff: Mentor

    To the OP: Homework problems need to be posted in the Homework & Coursework sections, not in the technical math sections. I have moved your thread. Please post any future questions in the appropriate section.
     
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