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Algebraic manipulation for easier integration

  1. Aug 8, 2009 #1
    1. The problem statement, all variables and given/known data

    I am trying to integrate x^2/(1+x^2 ) from 0 to 1.

    3. The attempt at a solution

    We recently worked on trig substitutions in class, but, rather than substituting x for tan(theta) I think there may be an easier way via algebraic manipulation. If I divide both numerator and denominator by the highest power of X in the denominator (x^2), then I get back out

    1/(1/x^2 +1)

    which is equal to x^2/2.

    Now, I can easily integrate 1/2 x^2 without trig substitution.

    My main question is: Is this a viable method for simplifying the integral, or must I go through trig substitution?
    Thanks
     
  2. jcsd
  3. Aug 8, 2009 #2

    rock.freak667

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    I don't think what you did made it simpler, but here is what I'd do


    [tex]\int_0 ^{1} \frac{x^2}{x^2+1}dx = \int_0 ^{1} \frac{x^2+1-1}{x^2+1}dx = \int_0 ^{1} \left( 1- \frac{1}{x^2+1} \right)dx[/tex]
     
  4. Aug 8, 2009 #3

    Hurkyl

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    :confused:
     
  5. Aug 8, 2009 #4
    Doesn't it? Algebra is not exactly my strong suit. 1/(1/x^2) is x^2 (the denominator of the denominator is the numerator), right?
     
  6. Aug 8, 2009 #5

    Hurkyl

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    This is true, but not relevant to the problem -- there's a "+1" in there, and you can't just ignore it.

    This is true:
    [tex]\frac{1}{\frac{1}{q}} = q[/tex]

    but you're working with
    [tex]\frac{1}{\frac{1}{q} + r}[/tex]

    which is not of the form of the left hand side.


    Incidentally, if you're ever unsure about an algebraic identity, you should do a sanity check -- try plugging in some numbers. If your algebraic manipulation is valid, then plugging in numbers should give equal results.

    (Though beware -- getting equal results doesn't prove your manipulation is valid)
     
  7. Aug 9, 2009 #6
    Thanks for the help. rock.freak667, I think what you did makes a lot more sense than what I did.
     
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