Conquering Limits: Solving the Integral from Negative to Positive Infinity

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Homework Help Overview

The discussion revolves around evaluating the improper integral from negative to positive infinity of the function dx/(x^2 + 6x + 12). The subject area pertains to integral calculus, specifically focusing on techniques for handling improper integrals and completing the square.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the method of completing the square to simplify the integral. There are questions about handling the improper nature of the integral and the approach to integrating the resulting expression. Some participants express concern about the original poster's engagement with the problem.

Discussion Status

The discussion is ongoing, with participants providing hints and guidance on how to approach the integral. There is an emphasis on the importance of the original poster demonstrating their understanding and progress in the problem-solving process.

Contextual Notes

Participants note the necessity for the original poster to show their work and where they are struggling, as this is crucial for receiving effective help. There is also mention of the integral being improper due to the limits of integration extending to infinity.

nyyfan0729
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The integral from negative infiniti to positive infiniti of dx/(x^2 + 6x + 12).
 
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Complete the square.
 
You have the integral
[tex] \begin{align*}<br /> \int \frac{1}{x^2+6x+12} dx<br /> \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*}<br /> \int \frac{1}{x^2+6x+12} dx &= \int \frac{1}{\left(x+3\right)^2 + 3} dx\\<br /> &= \int \frac{1}{u^2+3}du && \text{letting\ } u=x+3<br /> \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.
 
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I'm trying to take the improper integral but I just can't figure it out
 
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). We can't help you unless you show us what you have done, and where you're stuck.
 
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?
 
Hurkyl said:
BobMoretti: we really prefer it when helpers provide help, rather than simply doing most of the work for the poster.

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).
 
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