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Integration for BC Calculus

  1. Feb 12, 2006 #1
    The integral from negative infiniti to positive infiniti of dx/(x^2 + 6x + 12).
     
  2. jcsd
  3. Feb 12, 2006 #2

    arildno

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    Complete the square.
     
  4. Feb 12, 2006 #3
    You have the integral
    [tex]
    \begin{align*}
    \int \frac{1}{x^2+6x+12} dx
    \end{align*}
    [/tex]

    which as arildno suggested can be evaluated by completing the square. If you don't know what that means, you complete the square by realizing that [itex]x^2+6x+12 = (x+3)^2 +c[/itex] by means of the identity [itex](a+b)^2 = a^2 + 2ab+b^2[/itex].

    In this case, [itex]c=3[/itex]. So we can write
    [tex]
    \begin{align*}
    \int \frac{1}{x^2+6x+12} dx &= \int \frac{1}{\left(x+3\right)^2 + 3} dx\\
    &= \int \frac{1}{u^2+3}du && \text{letting\ } u=x+3
    \end{align*}
    [/tex]

    From this point you have an expression which you should be able to integrate by using the arctangent function.

    Now this is an improper integral, since the limits you gave are [itex]\pm \infty[/itex]. Do you know how to deal with those? You don't have to worry about a discontinuity since the demoninator is never 0.

    Hope that helps.
     
    Last edited: Feb 12, 2006
  5. Feb 12, 2006 #4
    I'm trying to take the improper integral but I just can't figure it out
     
  6. Feb 12, 2006 #5

    Hurkyl

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    BobMoretti: we really prefer it when helpers provide help, rather than simply doing most of the work for the poster.

    Now that being said...

    nyyfan0729: you've practically had the entire problem done for you, yet you show no signs that you've done anything (even if it was simply to process what's been shown). :grumpy: We can't help you unless you show us what you have done, and where you're stuck.
     
  7. Feb 12, 2006 #6

    arildno

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    nyyfan:
    It is important that you start playing about with the numbers given you, so that you develop an ability to recognize general patterns scantily masked in the individual cases.
    To give you one more hint:
    We have (since the limits of u are also plus and minus infinity):
    [tex]\int_{-\infty}^{\infty}\frac{du}{3+u^{2}}=\frac{1}{3}\int_{-\infty}^{\infty}\frac{du}{1+(\frac{u}{\sqrt{3}})^{2}}[/tex]
    Can you manage from here, then?
     
  8. Feb 14, 2006 #7
    Sorry, Hurkyl. I was mostly looking to spin my wheels with the cool embedded [itex]\LaTeX[/itex] stuff.

    Anyways, the subtler part of that problem involves dealing with the limits of integration, which I left to nyyfan.

    These boards seem very, very cool, and I hope that I'll be able to contribute in a constructive manner (without babying people).
     
    Last edited: Feb 14, 2006
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