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  1. #1
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    tough integral

    Could anyone help me with this monster?

    it's the integral 1/((x^4) + 4) dx

    Any help you could give would be Awesome!!!
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  2. #2
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    try a u substitution. that is, let u = (x^4) + 4 and then take the intergral of 1/u
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  3. #3
    Senior Member Pinkk's Avatar
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    Quote Originally Posted by o&apartyrock View Post
    try a u substitution. that is, let u = (x^4) + 4 and then take the intergral of 1/u
    Not possible since du=4x^{3}dx, and there is no 4x^{3} term in the integral.
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  4. #4
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    \int \frac{1}{x^4 + 4}dx = \int \frac{1}{(x^2 - 2x + 2)(x^2 + 2x + 2)}dx


    \frac{Ax + C}{x^2 - 2x + 2} + \frac{Bx + D}{x^2 + 2x + 2}


    Ax(x^2 + 2x + 2) + C(x^2 + 2x + 2) + Bx(x^2 - 2x + 2) + D(x^2 - 2x + 2) = 1


    Where C and D are 1/4,


    Ax(x^2 + 2x + 2) + 1/2x + Bx(x^2 - 2x + 2) = 0
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  5. #5
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    tried that

    I tried messing with a u sub but didn't get anywhere.

    My prof said that this problem uses most of the techniques, as far as integral calc goes. He then went on to say that once you get past clac 2, this problem is easy.

    Thanks prof! lol
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  6. #6
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    aahhh

    Quote Originally Posted by derfleurer View Post
    \int \frac{1}{x^4 + 4}dx = \int \frac{1}{(x^2 - 2x + 2)(x^2 + 2x + 2)}dx


    \frac{Ax + C}{x^2 - 2x + 2} + \frac{Bx + D}{x^2 + 2x + 2}


    Ax(x^2 + 2x + 2) + C(x^2 + 2x + 2) + Bx(x^2 - 2x + 2) + D(x^2 - 2x + 2) = 1


    Where C and D are 1/4,


    Ax(x^2 + 2x + 2) + 1/2x + Bx(x^2 - 2x + 2) = 0
    Hmmmm...

    Thanks that helps me get going. This doesn't look like it's going to be fun though. hehehe.
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  7. #7
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    Well I just checked wolfram, the answer they've got is

    \frac{1}{8}tan^{-1}(x + 1) - \frac{1}{8}tan^{-1}(1 - x) + \frac{1}{16}ln(x^2 + 2x + 2) - \frac{1}{16}ln(x^2 - 2x + 2)

    Edit: Working backwards, that's


    \frac{1}{8} \int \frac{1}{1 + (x + 1)^2}dx - \frac{1}{8} \int \frac{-1}{1 + (1 - x)^2}dx + \frac{1}{16} \int \frac{2x + 2}{x^2 + 2x + 2}dx - \frac{1}{16} \int \frac{2x - 2}{x^2 - 2x + 2}dx

    Edit: Where the denominators of the inverse tangents are just disguised forms of the other denominators.
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  8. #8
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    Quote Originally Posted by redepoch7 View Post
    Could anyone help me with this monster?

    it's the integral 1/((x^4) + 4) dx

    Any help you could give would be Awesome!!!
    \frac{4}{x^4 + 4} = \frac{x^2 + 2}{x^4 + 4} - \frac{x^2 - 2}{x^4 + 4}=\frac{1 + \frac{2}{x^2}}{(x - \frac{2}{x})^2 + 4}- \frac{1- \frac{2}{x^2}}{(x + \frac{2}{x})^2 - 4}. let y=x-\frac{2}{x} and z=x + \frac{2}{x}. then: 4 \int \frac{dx}{x^4 + 4} = \int \frac{dy}{y^2 + 4} - \int \frac{dz}{z^2 - 4}, and the rest is easy now.
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  9. #9
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    Quote Originally Posted by NonCommAlg View Post
    \frac{4}{x^4 + 4} = \frac{x^2 + 2}{x^4 + 4} - \frac{x^2 - 2}{x^4 + 4}=\frac{1 + \frac{2}{x^2}}{(x - \frac{2}{x})^2 + 4}- \frac{1- \frac{2}{x^2}}{(x + \frac{2}{x})^2 - 4}. let y=x-\frac{2}{x} and z=x + \frac{2}{x}. then: 4 \int \frac{dx}{x^4 + 4} = \int \frac{dy}{y^2 + 4} - \int \frac{dz}{z^2 - 4}, and the rest is easy now.
    sorry but where did the 4 in the numerator of part one come from? Does it matter? Can you just factor that out?

    ex. \frac{4}{x^4 + 4}
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  10. #10
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by redepoch7 View Post
    sorry but where did the 4 in the numerator of part one come from? Does it matter? Can you just factor that out?

    ex. \frac{4}{x^4 + 4}
    Since you were dealing with \frac{1}{x^4+1}, follow NonCommAlg's suggestion, but then multiply your answer by \frac{1}{4}.
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