Conquering Limits: Solving the Integral from Negative to Positive Infinity

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SUMMARY

The integral from negative to positive infinity of the function dx/(x^2 + 6x + 12) can be solved by completing the square, resulting in the expression (x+3)^2 + 3. This transformation allows the integral to be rewritten as ∫(1/(u^2 + 3)) du, where u = x + 3. The improper integral can be evaluated using the arctangent function, specifically through the identity ∫(du/(3 + u^2)) = (1/3)∫(du/(1 + (u/√3)^2)). Understanding how to handle improper integrals is crucial for solving this problem.

PREREQUISITES
  • Understanding of integral calculus
  • Familiarity with completing the square technique
  • Knowledge of improper integrals
  • Proficiency in using the arctangent function for integration
NEXT STEPS
  • Study the method of completing the square in polynomial expressions
  • Learn about improper integrals and their evaluation techniques
  • Explore the properties and applications of the arctangent function in integration
  • Practice solving similar integrals involving limits of integration from negative to positive infinity
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Students and educators in mathematics, particularly those focused on calculus, integral evaluation, and advanced integration techniques.

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
<br /> \begin{align*}<br /> \int \frac{1}{x^2+6x+12} dx<br /> \end{align*}<br />

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 x^2+6x+12 = (x+3)^2 +c by means of the identity (a+b)^2 = a^2 + 2ab+b^2.

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

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 \pm \infty. 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):
\int_{-\infty}^{\infty}\frac{du}{3+u^{2}}=\frac{1}{3}\int_{-\infty}^{\infty}\frac{du}{1+(\frac{u}{\sqrt{3}})^{2}}
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 \LaTeX 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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