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     \frac {1}{1+x^5}

    I thought of substituting u = 1 + x^5, or u = x^5, but I can't seem to get either way to work. How do I go about integrating this?
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    Is that definitely the question because look at the solution: http://www.wolframalpha.com/input/?i=\int+1%2F%281%2Bx^5%29
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    (Original post by Sasukekun)
     \frac {1}{1+x^5}

    I thought of substituting u = 1 + x^5, or u = x^5, but I can't seem to get either way to work. How do I go about integrating this?
    http://www.wolframalpha.com/input/?i...1%2Bx%5E5%29dx
    :nope: Let it go...
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    With difficulty. Unless there's some kind of elaborate trig substitution you can use, I'd recommend writing
    1+x^5 = (1+x)(1+\omega x)(1+\omega^2 x)(1+\omega^3 x)(1+\omega^4 x)
    where \omega = e^{\frac{2\pi i}{5}} is a fifth root of unity, and then do some crazy partial fractions. You'll end up with a load of complex logarithms which you can probably simplify a bit.
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    Dependant upon the limits of integration this is really nice. However if you are dealing with an indefinite integral of the form 1/1+x^n things get non trivial pretty fast.
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    *If this is C1 integration, then split the expression up, so it'll look like: (1/1) + (1/x^5) = 1 + x^-5
    then f(x) = x - ((x^-4)/4) + c

    I have a feeling that you're not doing c1 integration... But hope it helps anyway!
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    (Original post by sweetascandy)
    *If this is C1 integration, then split the expression up, so it'll look like: (1/1) + (1/x^5) = 1 + x^-5
    then f(x) = x - ((x^-4)/4) + c

    I have a feeling that you're not doing c1 integration... But hope it helps anyway!
    \frac{a}{b+c} \not \equiv \frac{a}{b} + \frac{a}{c}
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    (Original post by Sasukekun)
     \frac {1}{1+x^5}

    I thought of substituting u = 1 + x^5, or u = x^5, but I can't seem to get either way to work. How do I go about integrating this?
    split it into partial fractions and do integration by substitution, or you could do by parts by having u as (1+x^5)^-1 and v' as 1,

    EDIT:actually by parts would probably be really hard, and might not give you a solution
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    \dfrac{1}{1+x^5}\equiv\dfrac{1}{  1+(x^{2.5})^2}

    x^{2.5}=tan(u)\Leftrightarrow 2.5x^{1.5}\dfrac{dx}{du}=sec^2(u  )\Leftrightarrow dx=\dfrac{sec^2(u)}{2.5x^{1.5}}

    \displaystyle\int\frac{sec^2(u)}  {2.5x^{1.5}(1+tan^2(u))}}\ du=\frac{2}{5}\int x^{-1.5}\ du



    can you finish it off? :holmes:

    Spoiler:
    Show
    i'm aware the first statement is false
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    (Original post by sweetascandy)
    *If this is C1 integration, then split the expression up, so it'll look like: (1/1) + (1/x^5) = 1 + x^-5
    then f(x) = x - ((x^-4)/4) + c

    I have a feeling that you're not doing c1 integration... But hope it helps anyway!
    lmao
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    If you work backwards you get y = (1/5x^4)ln(1+x^5)
    Not sure if this works though. Check by differentiating y of course, if you get 1/(1+x^5) then y is right.
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    Yeah, thanks all guys. It is hard, but the original question was a double integral, and I didn't realise I could work around what was asked initially.
 
 
 
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