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Math Help - Integral (1+x^2)^{-2}

  1. #1
    Junior Member gusztav's Avatar
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    Integral (1+x^2)^{-2}

    Hello, I'm trying to solve this integral

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

    and would be really grateful for any help!
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  2. #2
    Lord of certain Rings
    Isomorphism's Avatar
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    Quote Originally Posted by gusztav View Post
    Hello, I'm trying to solve this integral

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

    and would be really grateful for any help!
    Did you try the sub, x = \tan \theta?
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  3. #3
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    Why dont we generalize this problem :  \int \frac{dx}{(1+ x^2)^n}

    Let I_{n} be the integral ,
     I_{n} = \int \frac{1 + x^2 - x^2}{(1+x^2)^n} ~dx

     = \int \frac{1}{(1+x^2)^{n-1}} - \frac{x^2}{(1+x^2)^n} ~dx

     = I_{n-1} - x \frac{-1}{2(n-1)(1+x^2)^{n-1}} + \frac{-1}{2(n-1)} \int \frac{dx}{(1+x^2)^{n-1}} ~~~ integration~by~parts

     = I_{n-1} + \frac{x}{2(n-1)(1+x^2)^{n-1}} - \frac{1}{2(n-1)} I_{n-1}<br />

    Therefore ,  I_{n} = \frac{2n-3}{2n-2} I_{n-1} + \frac{x}{2(n-1)(1+x^2)^{n-1}}

    for n = 2 ,  I_2 = \frac{1}{2} \tan^{-1}(x) + \frac{x}{2(1+x^2)^{1}}
    Last edited by simplependulum; June 11th 2009 at 02:57 AM.
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  4. #4
    MHF Contributor chisigma's Avatar
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    Is...

    \frac{1}{(1+x^{2})^{2}} = \frac{1 + x^{2} - x^{2}}{(1+x^{2})^{2}} = \frac{1}{1+x^{2}} - x\cdot \frac{x}{(1+x^{2})^{2}}

    ... so that...

    \int \frac{dx}{(1+x^{2})^{2}} = \int \frac{dx}{1+x^{2}} + \int x\cdot \frac{x\cdot dx}{(1+x^{2})^{2}}

    The first integral is well known, the second can be solved by parts...

    Kind regards

    \chi \sigma
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