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Math Help - Indefinite integral of (x+(1/x))^4

  1. #1
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    Wink Indefinite integral of (x+(1/x))^4

    Hello, I'm Fahad and I'm new here
    I'm here to get help with simple math problems for my IGCSES
    Plese help me.
    The following math is from the binomial expansions part of my syllabus:

    ⌠ (x+(1/x))^4

    Please show your working and thanks for the help.
    Also, please refer to the techniques of integration you are using.
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  2. #2
    Senior Member MaxJasper's Avatar
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    Lightbulb Re: Indefinite integral of (x+(1/x))^4

    \int \left(x+\frac{1}{x}\right)^4 \, dx = \int \left(6+\frac{1}{x^4}+\frac{4}{x^2}+4 x^2+x^4\right) \, dx
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  3. #3
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    Re: Indefinite integral of (x+(1/x))^4

    Please show your working throughout and please simplify
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  4. #4
    MHF Contributor MarkFL's Avatar
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    Re: Indefinite integral of (x+(1/x))^4

    We are given:

    \int \left(x+\frac{1}{x} \right)^4\,dx

    Using the binomial theorem on the integrand, we find:

    \left(x+\frac{1}{x} \right)^4=\left(x+x^{-1} \right)^4=\sum_{k=0}^4{4 \choose k}x^{4-k}x^{-k}

    After simplification:

    x^4+4x^2+6+4x^{-2}+x^{-4}

    So then we have:

    I=\int x^4+4x^2+6+4x^{-2}+x^{-4}\,dx

    We will integrate term by term, using the power rule for integration \int k\cdot x^n\,dx=\frac{k}{n+1}x^{n+1}+C

    where k is a constant and n\ne-1

    So we have:

    I=\frac{1}{5}x^5+\frac{4}{3}x^3+6x-4x^{-1}-\frac{1}{3}x^{-3}+C
    Last edited by MarkFL; September 14th 2012 at 12:28 AM.
    Thanks from fbarawr
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