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Thread: difficult integration: int [ 1 / (1 + x^4) ] dx

  1. #1
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    difficult integration: int [ 1 / (1 + x^4) ] dx

    integrate:

    dx/ (1 + x[sup:2fussihg]4[/sup:2fussihg])

    I've tried by parts using u = 1 / (1 + x[sup:2fussihg]4[/sup:2fussihg]) and dv = dx, but that only increased the power of the function

    I also tried using the substitution x = SQRT(tant), with dx = sec[sup:2fussihg]2[/sup:2fussihg](t) / (2SQRT(tant)). This seemed to simplify it a bit but I still got stuck integrating.

  2. #2
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    Re: difficult integration

    Have you considered factoring?

    [tex]x^{4}+1 = (x^{2}+\sqrt{2}x+1)(x^{2}-\sqrt{2}x+1)[/tex]

    That may lead somewhere if you are courageous.
    "Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.

  3. #3
    Elite Member stapel's Avatar
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    Quote Originally Posted by dts5044
    integrate: dx/ (1 + x[sup:wq5pahuc]4[/sup:wq5pahuc])
    The Integrator would appear to suggest that factorization, as nasty as that will be, is the way to go.

    Eliz.

  4. #4
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    Re: difficult integration: int [ 1 / (1 + x^4) ] dx

    ...and the associated partial fractions are no additional picnic.

    This is an excellent argument for computer-based examination of such things. Is all this pain really necessary?
    "Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.

  5. #5
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    Re: difficult integration: int [ 1 / (1 + x^4) ] dx

    1/(1+x^4) = 1/[1-(-x^4)} = Sum [(-1)^n*x^(4n)], n goes from 0 to infinity.

    integral of [1/(1+x^4)]dx = Sum[(-1)^n*x^(4n+1)]/(4n+1), n goes from 0 to infinity.

    I'll leave as an exercise to find the interval of convergence.

  6. #6
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    Re: difficult integration: int [ 1 / (1 + x^4) ] dx

    Hello, dts5044!

    Want to see a truly far-out solution?


    [tex]\int\frac{dx}{1 + x^4}[/tex]

    [tex]\text{We have: }\;\frac{1}{x^4 + 1} \;=\;\frac{1}{2}\bigg[\frac{x^2 + 1 - x^2 + 1}{x^4+1}\bigg] \;=\;\frac{1}{2}\bigg[\frac{x^2+1}{x^4+1} - \frac{x^2-1}{x^4+1}\bigg][/tex]

    [tex]\text{Divide top and bottom by }x^2\!:\;\;\frac{1}{2}\left[\frac{\frac{x^2+1}{x^2}}{\frac{x^4+1}{x^2}} - \frac{\frac{x^2-1}{x^2}}{\frac{x^4+1}{x^2}} \right][/tex]

    [tex]\text{And we will integrate: }\;\frac{1}{2}\left[\int\frac{\left(1 + \frac{1}{x^2}\right)\,dx}{x^2 + \frac{1}{x^2}} - \int\frac{\left(1-\frac{1}{x^2}\right)\,dx}{x^2 + \frac{1}{x^2}}\right][/tex] .[1]

    You think that's weird? . . . Just wait!


    [tex]\text{Let }\,u \:=\:x - \frac{1}{x}\quad\Rightarrow\quad du \:=\:\left(1 + \frac{1}{x^2}\right)\,dx[/tex]

    [tex]\text{Let }\,v \;=\:x + \frac{1}{x}\quad\rightarrow\quad dv \:=\:\left(1 - \frac{1}{x^2}\right)\,dx[/tex]

    . . [tex]u^2\:=\:x^2 - 2 + \frac{1}{x^2}\quad\Rightarrow\quad u^2+2 \:=\:x^2 + \frac{1}{x^2}[/tex]

    . . [tex]v^2 \:=\:x^2 + 2 + \frac{1}{x^2}\quad\Rightarrow\quad v^2-2 \:=\:x^2 + \frac{1}{x^2}[/tex]


    Substitute into [1]:

    . . . . [tex]\frac{1}{2}\left[\int\frac{du}{u^2 + 2} - \int\frac{dv}{v^2-2}\right][/tex]


    Just integrate and back-substitute . . .


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    I wish I could take credit for this innovative approach,
    . . but it was posted in February 2007 by a tutor named "commutative".
    (And I can't locate the site where it appeared.)

    I'm the other of the two guys who "do" homework.

  7. #7
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    Re: difficult integration: int [ 1 / (1 + x^4) ] dx

    If complex number is allowable then:

    [tex]x^4 + 1 = (x^2)^2 - (i)^2 = (x^2 - i)\cdot (x^2 + i)[/tex]

    Break into partial fractions and then standard integration.
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

  8. #8
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    Re: difficult integration: int [ 1 / (1 + x^4) ] dx

    I have not posted on this topic because it is a perfect example of what I think is wrong the mathematics education today.
    The question does not ask if the integral exists.
    That is a theoretical question that is important if we ought to continue.
    If the function is integrable then any reasonable CAS will find the value of integrals.
    How one finds the so-called ‘primitive’ (antiderivative) is really moot!
    So my question is: “why are you concerned?”
    “A professor is someone who talks in someone else’s sleep”
    W.H. Auden

  9. #9
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    Re: difficult integration: int [ 1 / (1 + x^4) ] dx

    I wondered how long it would take. I was baiting you.
    "Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.

  10. #10
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    Re: difficult integration: int [ 1 / (1 + x^4) ] dx

    'Commutative' is a very learned contributor on SOS Math

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