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A difficult integral, perhaps involving trig substitutions?

  1. Jun 22, 2009 #1
    1. The problem statement, all variables and given/known data

    The integral of 1/(1+x^6) dx

    3. The attempt at a solution

    This problem was given as a "challenge/bonus problem" on the first day of Calc 2, earlier today. Considering we were going over trig substitutions, I suppose there is probably some trig trick to it. Doing it by parts isn't helping... I just get a longer more messy expression to integrate. I also tried a u substitution. I set u = x^(3) to make the integral seem like I could pull out an arctan, But I keep ending up only making the situation worse.

    I've typed in the expression into an online integral calculator so I know what the answer is, but I'm confused on how to go about tackling this problem. Any help would be greatly appreciated.
     
    Last edited: Jun 22, 2009
  2. jcsd
  3. Jun 22, 2009 #2

    rock.freak667

    User Avatar
    Homework Helper

    What answer did it give?
     
  4. Jun 22, 2009 #3

    Mark44

    Staff: Mentor

    The denominator can be factored into (x2 + 1)(x4 - x2 + 1), and the quartic factor can be factored into (x2 - 1/2 - i/2*sqrt(3))(x2 - 1/2 + i/2*sqrt(3)). At that point you could break up the integral using the technique of partial fractions.
     
  5. Jun 23, 2009 #4
    Thanks mark, can you give me the name of the technique used to factor out the quartic function, or give a little explanation yourself? I can see that it works but I'm curious as how one would go about splitting up a quartic function using imaginary numbers. I have yet to come across this technique in any of my classes. The answer I got for the integral is quite long so I'm not going to write it out but I used the integral calculator found at http://integrals.wolfram.com/index.jsp

    Thanks again
     
  6. Jun 23, 2009 #5

    Mark44

    Staff: Mentor

    All I used to factor the quartic was the quadratic formula, since this particular quartic was quadratic in form, with a 4th-degree term, a 2nd-degree term, and a constant term.

    If you substitute u = x2, the quartic part becomes u2 - u + 1 = 0, and then use the quadratic formula to solve for u (= x2).
     
  7. Jun 25, 2009 #6
    Okay, I sat down today and tried to solve the thing and still can't get it. When I do the partial fractions method I just end up with long ridiculous equations with imaginary numbers in them that don't cancel out. Am I supposed to split up each of the terms again? Or is there a whole different way to approach this problem? I was talking about it with someone who knows how to do it but won't tell me and he looked confused when I mentioned imaginary numbers.

    Also just for reference the answer the online integral calculator gave was...

    (sqrt(3)*log(x^2+sqrt(3)*x+1)-sqrt(3)*log(x^2-sqrt(3)*x+1)+2*atan(2*x+sqrt(3))+2*atan(2*x-sqrt(3))+4*atan(x))/12
     
  8. Jun 25, 2009 #7
    Notice that [tex] x^6 + 1 = (x^2 +1)(x^4 - x^2 + 1) [/tex].

    Since [tex] x^4 - x^2 + 1 = (x^2 + 1)^2 - 3x^2 = (x^2 + 1 - \sqrt(3)x) (x^2 + 1 + \sqrt(3)x) [/tex] **Note that the x is not under the sqrt sign**

    And so [tex] x^6 + 1 = (x^2 +1)(x^2 + 1 - \sqrt(3)x) (x^2 + 1 + \sqrt(3)x) [/tex].

    You can now use partial fractions.
     
  9. Jun 25, 2009 #8
    [tex]
    x^4 - x^2 + 1 = (x^2 + 1)^2 - 3x^2
    [/tex]

    Thanks for the quick reply! I see that this works... is this a common algebraic maneuver for factoring quartic functions? I have never seen anything like this up to this point.
     
  10. Jun 25, 2009 #9
    I encountered this question a couple months ago, and after a couple frustrating days I finally realized what the trick was.

    It was the first time I've also seen something like it :)
     
  11. Jun 25, 2009 #10
    You complete the square and obtain a difference of squares, so you can easily factor from there. If I spot a root immediately, then I use synthetic division; if not, then I use this method. But that's just me :)
     
  12. Jun 25, 2009 #11
    :frown:

    I keep coming back to this trying to solve it and just can't get it. Am I supposed to set up partial fractions over the these three terms,

    [tex] x^6 + 1 = (x^2 +1)(x^2 + 1 - \sqrt(3)x) (x^2 + 1 + \sqrt(3)x) [/tex]

    because even if I did, I wouldn't know how to integrate A/(x^2+1+3^(1/2)*x)... but when I try to split that up using the quadratic formula I get another imaginary number. Am I just supposed to use partial fractions over those 3 terms and then just use integration by parts to solve for some constant over a quadratic?
     
  13. Jun 25, 2009 #12
    When doing partial fractions with quadratics in the denominator, you must use Ax+B, Cx+D, and Ex+F, not A, B, and C. After you do partial fractions, you can algebraically manipulate each expression. For instance, let's say you have to integrate 2xdx/(x^2+sqrt(3)*x+1). You can rewrite this as the integral of (2x+sqrt(3)-sqrt(3))/(x^2+sqrt(3)*x+1). You can then split this up into two integrals; for the first one, you can use substitution, and for the second you will have to manipulate it again by completing the square in the denominator and then using a trig-sub (your answer will involve arctan).
     
  14. Jun 25, 2009 #13

    zcd

    User Avatar

    It would be Ax + B and Cx + D and etc for the numerators, not just a single A B or C constant. If you end up with just a constant, like A/(x^2+1+3^(1/2)*x), then you could complete the square and integrate for an arctan solution.
     
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