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Math Help - Very simpl integral

  1. #1
    Super Member dhiab's Avatar
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    Very simpl integral

    Calculate :
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  2. #2
    Senior Member DeMath's Avatar
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    \int {\frac{{dx}}{{{x^3} + x}}}  = \int {\frac{{{x^2} + 1 - {x^2}}}{{x\left( {{x^2} + 1} \right)}}dx}  = \int {\frac{{dx}}{x}}  - \int {\frac{x}{{{x^2} + 1}}dx}  =

    = \ln \left| x \right| - \frac{1}{2}\int {\frac{{2x}}{{{x^2} + 1}}} dx = \ln \left| x \right| - \frac{1}{2}\int {\frac{{d\left( {{x^2} + 1} \right)}}{{{x^2} + 1}}}  =

    = \ln \left| x \right| - \frac{1}{2}\ln \left( {{x^2} + 1} \right) + C.<br />
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  3. #3
    Super Member flyingsquirrel's Avatar
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    Hi,
    Quote Originally Posted by dhiab View Post
    Calculate :
    Substitute x=1/t or use a partial fraction decomposition.
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  4. #4
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by dhiab View Post
    Calculate :
    \frac{1}{x^3+x}=\frac{1}{x}+\frac{-x}{x^2+1}

    \int \frac{1}{x} dx + \int \frac{-x}{x^2+1} dx

    the first is ln(x) the second by substitute u=x^2+1 du=2x dx

    it is clear right
    Last edited by mr fantastic; July 17th 2009 at 08:20 PM. Reason: Fixed the tag in second line of latex
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  5. #5
    Super Member dhiab's Avatar
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    Thanks and anathor solution

    Thanks for all : Demath , Flyinquirrel, Amer
    I'have anathor solution :
    Attached Thumbnails Attached Thumbnails Very simpl integral-18.jpg  
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