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U-sub integral: Not sure where i went wrong.

  1. Jan 21, 2013 #1
    1. The problem statement, all variables and given/known data

    Im not sure what i did wrong, every operation i took to evaluate the integral seems valid. I took pictures of the integral evaluated two different ways, producing two different answers. What bugs me is even if i evaluate (as i did) from say 0 to 1 the numbers are different, so they are not equivalent. I dont see any errors in the first one, but the text book lists the latter as a correct answer.

    The incorrect way:
    2QfhdVs.jpg
    Correct:
    24E2jat.jpg
    2. Relevant equations



    3. The attempt at a solution
    The incorrect way:
    <a href="[PLAIN]http://imgur.com/2QfhdVs"><img [Broken] src="http://i.imgur.com/2QfhdVs.jpg?1" title="Hosted by imgur.com" alt="" /></a>[/PLAIN]

    Correct:
    <a href="[PLAIN]http://imgur.com/24E2jat"><img [Broken] src="http://i.imgur.com/24E2jat.jpg" title="Hosted by imgur.com" alt="" /></a>[/PLAIN]
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Jan 21, 2013 #2

    Dick

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    Saying u/(2+u^2-2u) equals u/2+u^2/u-2u/u is seriously bad algebra. (a+b)/c=a/c+b/c. c/(a+b) IS NOT equal to c/a+c/b. Try some numbers.
     
  4. Jan 21, 2013 #3

    SammyS

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    Your error is in doing basic algebra with a rational expression.

    [itex]\displaystyle \frac{u}{u^2-2u+2}\ne\frac{u}{u^2}-\frac{u}{2u}+\frac{u}{2}[/itex]

    After all, is [itex]\displaystyle \ \ \frac{1}{1+2+3}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}\ ? \ \ [/itex]
     
    Last edited by a moderator: May 6, 2017
  5. Jan 21, 2013 #4
    oh christ. im dumb.

    Apologies.
     
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