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Improper Integral

  • Thread starter tahayassen
  • Start date
  • #1
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[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.
 

Answers and Replies

  • #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:
  • #3
269
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Have you tried the following?
[tex]x^6+9 = (x^3-3)(x^3+3)[/tex]?
Then:
[tex]\frac{1}{(x^3-3)(x^3+3)} = \frac{1}{2}(\frac{1}{x^3-3}+\frac{1}{x^3+3})[/tex]
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.
 
  • #4
Yes, sorry, I edited my original post :)...
Spoke too soon. How about the alternative?
 
  • #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.
 
  • #6
HallsofIvy
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[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 by a moderator:
  • #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.
 
  • #8
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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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