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Pattern recognition on integrals questions

  1. Jul 3, 2013 #1
    Hello users,
    I would like to know when do you use pattern recognition over integrals
    Someone told me it was that
    For example the integral below

    I would like to know the procedure to rewrite the numerators as (2x-2) + 3
    Where does the 3 come from?
    I would really appreciate

    Thanks in advance!
     

    Attached Files:

  2. jcsd
  3. Jul 3, 2013 #2

    Ackbach

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    Gold Member

    Pattern recognition is at the heart of mathematics no matter which branch of math you're dealing with. In this case, you can complete the square in the denominator. Rewriting the numerator the way you've mentioned then makes the squared term in the denominator appear in the numerator. Then you can try a change of variables. You get:
    [tex]\int \frac{2x+1}{(x-1)^{2}-3}\,dx = \int \frac{(2x-2)+3}{(x-1)^{2}-3}\,dx
    =\int \frac{2(x-1)+3}{(x-1)^{2}-3} \, dx.[/tex]
    Substitute [itex]u=x-1[/itex] to simplify.
     
  4. Jul 3, 2013 #3

    Makes sense,, thank you,, but could you explain me how did you get the number 3 in the numerator??
    I know that the integral is in the form f'(x)/f(x),, but what do you do with the constant?
    Im a little confused
     
    Last edited: Jul 3, 2013
  5. Jul 3, 2013 #4

    Ackbach

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    Gold Member

    Sure. You can't actually change the numerator, can you? You can rewrite it to look a little different, but that's all. The [itex]3[/itex] in the numerator is the number you must add to [itex]-2[/itex] in order to ensure that you end up with the original [itex]+1[/itex]. So the thinking goes this way: I want a [itex]-2[/itex] there, but I have a [itex]+1[/itex]. How could I get a [itex]-2[/itex]? By rewriting [itex]1=-2+3[/itex]. So that's where the [itex]3[/itex] comes from.
     
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