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Math Help - Partial fractions

  1. #1
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    Partial fractions

    does anyone now how to integrate this by partial frac?
    \int\frac{1}{(t^2+1)^3}

    using quickmath wouldnt work
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  2. #2
    Senior Member tukeywilliams's Avatar
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    Use the following form:  \frac{A_{1}x + B_1}{ax^2+bx+c} + \frac{A_{2}x + B_2}{(ax^2+bx+c)^2} + \frac{A_{3}x + B_3}{(ax^2+bx+c)^3}  . In this case its  x^2 +1,  (x^2+1)^2 , and  (x^2+1)^3 . Then solve for the coefficients.

    So:  \frac{1}{(x^2+1)^3} =  \frac{A_{1}x + B_1}{x^2+1} + \frac{A_{2}x + B_2}{(x^2+1)^2} + \frac{A_{3}x + B_3}{(x^2+1)^3}
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  3. #3
    MHF Contributor red_dog's Avatar
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    \displaystyle \frac{1}{(t^2+1)^3} is already a partial fraction.
    You have to integrate by parts.
    There is a recurrence relation between I_n and I_{n-1}.
    \displaystyle I_n=\int\frac{1}{(x^2+a^2)^n}dx=\frac{1}{a^2}\left[\frac{x}{2(n-1)(x^2+a^2)^{n-1}}+\frac{2n-3}{2(n-1)}I_{n-1}\right],n\geq 2
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  4. #4
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    thanks...
    hi can you show me how integrating by parts is done?
    just the let u's and the let v's
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  5. #5
    MHF Contributor red_dog's Avatar
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    \displaystyle I_n=\frac{1}{a^2}\int\frac{x^2+a^2-x^2}{(x^2+a^2)^n}dx=\frac{1}{a^2}\left[I_{n-1}-\int\frac{x^2}{(x^2+a^2)^n}dx\right]=
    \displaystyle =\frac{1}{a^2}\left[I_{n-1}+\int x\left(\frac{1}{2(n-1)(x^2+a^2)^{n-1}}\right)'dx\right]
    and you can continue from here.
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  6. #6
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    i cant follow the last part
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  7. #7
    Math Engineering Student
    Krizalid's Avatar
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    Use

    \color{blue}\int {\left( {x^2 + a^2 } \right)^m ~dx} = \frac{{x\left( {x^2 + a^2 } \right)^m }}<br />
{{2m + 1}} + \frac{{2a^2 m}}<br />
{{2m + 1}}\int {\left( {x^2 + a^2 } \right)^{m - 1} ~dx},~\forall m\ne-\frac12

    Then if you set \color{red}m=-n will yield red_dog's formula.

    This reductions formulas are proved using integration by parts.
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  8. #8
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    galactus's Avatar
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    Hey Adam:

    Perhaps try it this way.


    \int\frac{1}{(x^{2}+1)^{3}}dx

    Let u=tan^{-1}(x), \;\ x=tan(u), \;\ dx=sec^{2}(u)du

    Make the subs and we get:

    \int\frac{sec^{2}(u)}{(1+tan^{2}(u))^{3}}du

    This equals:

    \int{cos^{4}(u)}du

    Handy reduction formula:

    \int{cos^{n}(u)}du-\frac{cos^{n-1}(u)sin(u)}{n}+\frac{n-1}{n}\int{cos^{n-2}}du

    EDIT: I'm sorry, Adam, I seen you had to use PF. Oh well, there it is anyway.
    Last edited by galactus; July 29th 2007 at 09:13 AM.
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