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Improper Integral

  1. Feb 2, 2013 #1
    [tex]\begin{array}{l}
    \int\limits_{ - \infty }^\infty {\frac{{{x^2}}}{{{x^6} + 9}}} \\
    = \int\limits_{ - \infty }^0 {\frac{{{x^2}}}{{{x^6} + 9}}} + \int\limits_0^\infty {\frac{{{x^2}}}{{{x^6} + 9}}} \\
    = \mathop {\lim }\limits_{t \to - \infty } \int\limits_t^0 {\frac{{{x^2}}}{{{x^6} + 9}}} + \mathop {\lim }\limits_{m \to \infty } \int\limits_0^m {\frac{{{x^2}}}{{{x^6} + 9}}}
    \end{array}[/tex]

    I had this problem on a test yesterday and I couldn't solve it. I tried to reduce the denominator to irreducible quadratic factors and then use partial fractions, but it was impossible. I couldn't see an obvious substitution, but apparently you were supposed to solve it by substitution.
     
  2. jcsd
  3. Feb 2, 2013 #2
    Have you tried the following?
    [tex]x^6+9 = (x^2+\sqrt[3]{3})(x^4-x^2\sqrt[3]{3}+3\sqrt[3]{3})[/tex]?
    Then via Fractional decomposition?
    (Edited!, mistook +9, for -9)
     
    Last edited: Feb 2, 2013
  4. Feb 2, 2013 #3
    HOW DID YOU GET THOSE FACTORS? I spent ages trying to factor it!

    Edit: Wait a second. You can't apply partial fractions to cubic factors.
     
  5. Feb 2, 2013 #4
    Yes, sorry, I edited my original post :)...
    Spoke too soon. How about the alternative?
     
  6. Feb 2, 2013 #5
    Okay, here's a way:
    Try dividing the denominator by 9;
    You should get:
    [tex]\frac{1}{9}\frac{x^2}{\frac{x^6}{9}+1}[/tex]
    Then set z^2(!) = x^6/9;
    By differentiation:
    [tex]2zdz=\frac{6x^5}{9}dx[/tex]
    Substitute, and integrate as necessary.
     
  7. Feb 2, 2013 #6

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    [itex](x^3+ 3)(x^3- 3)= x^6- 9[/itex], not [itex]x^6+ 9[/itex], which is why daniel akkerma changed his factorization. (But too late, his mistake will "live in infamy"!):tongue:

    Personally, I see no reason to factor [itex]x^6+ 9[/itex] at all. Instead, with that "[itex]x^2[/itex]" in the numerator, let [itex]u= x^3[/itex] so that [itex]du= 3x^2dx[/itex] and [itex](1/3)du= x^2dx[/itex] so the integral becomes
    [tex]\frac{1}{3}\int\frac{du}{u^2+ 6}[/tex]
    which can be integrated as an arctangent.

    (edit- That appears to be what daniel akkerma is suggesting the his post just before mine- that was posted while I was still typing.)
     
    Last edited: Feb 2, 2013
  8. Feb 2, 2013 #7
    Yeah, I am really sorry about, Ivy & Taha.
    For some reason I descried +9 as -9... Mysterious are the ways of the brain, etc, etc, etc :D...
    Sorry!
    Excellent choice of substitution from you, btw. I still like mine via z^2 though; Feels more direct.
     
  9. Feb 2, 2013 #8

    Mark44

    Staff: Mentor

    One thing that no one mentioned yet is that since the integrand is an even function, it suffices to find ## \int_0^{\infty} f(x) dx## and double it.
     
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