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Math Help - Integrating by Parts

  1. #1
    Member Em Yeu Anh's Avatar
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    Red face Integrating by Parts

    \int{x(arctanx)^2} dx

    No work to show so far, not sure which substitutions would be the best to make here.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Em Yeu Anh View Post
    \int{x(arctanx)^2} dx

    No work to show so far, not sure which substitutions would be the best to make here.
    integration by parts: u = (\arctan x)^2,~dv = x~dx

    think you can take it from here?
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  3. #3
    Member Em Yeu Anh's Avatar
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    Quote Originally Posted by Jhevon View Post
    integration by parts: u = (\arctan x)^2,~dv = x~dx

    think you can take it from here?
    If I have done this correctly,
    du=2(\arctan x)\frac{1}{1+x^2}dx, ~v=\frac{1}{2}x^2

    the integral is then \frac{1}{2}x^2(arctan x)^2 - \int \frac{1}{2}x^22(\arctan x)\frac{1}{1-x^2}dx

    I am then stuck on what to do with that integral..should I make the substitution u = arctan x and then do parts again?
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  4. #4
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    \frac{1}{2}x^2(\arctan x)^2 - \int x^2(\arctan x)\frac{1}{1+x^2}dx

    \frac{1}{2}x^2(\arctan x)^2 - \int \left(1 - \frac{1}{1+x^2}\right)\arctan{x} \, dx

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

    continue ...
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  5. #5
    Super Member General's Avatar
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    =Em Yeu Anh;440924]If I have done this correctly,
    the integral is then \frac{1}{2}x^2(arctan x)^2 - \int \frac{1}{2}x^22(\arctan x)\frac{1}{1{\color{red}-}x^2}dx
    The red one should be "+" not "-".
    Also the 2`s will cancel.
    and the integral will be:
    \int tan^{-1}(x) \frac{x^2}{x^2+1} dx
    Clearly: \frac{x^2}{x^2+1}=1-\frac{1}{x^2+1} <--- Do you know why?
    Substituting this in the integral:
    \int tan^{-1}(x) (1-\frac{1}{x^2+1}) dx
    =\int tan^{-1}(x)dx-\int \frac{tan^{-1}(x)}{x^2+1}dx
    This first one can be solved using integration by parts.
    The second one can be solved by using a U-Substitution.
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