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Thread: Finding integral of x tan^(-1) x dx

  1. #1
    Member helloying's Avatar
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    Finding integral of x tan^(-1) x dx

    Find $\displaystyle \frac{d}{dx} [tan^{-1} (x^2)] $ I found it to be $\displaystyle \frac{1}{1+x^4}$

    Hence find $\displaystyle \int x tan^{-1} (x^2) dx $

    i do integration by parts with u=$\displaystyle tan^{-1} (x^2)$
    $\displaystyle \frac{du}{dx}=\frac{1}{x^4+1}$
    v=$\displaystyle \frac{x^2}{2}$

    Then i got $\displaystyle 0.5x^2 tan^{-1} (x^2) - \int\frac{x^2}{2x^4 +2}$

    i got stuck here. pls help!
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  2. #2
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    Quote Originally Posted by helloying View Post
    Find $\displaystyle \frac{d}{dx} [tan^{-1} (x^2)] $ I found it to be $\displaystyle \frac{1}{1+x^4}$



    This is wrong. Read and apply correctly the chain rule and try again.

    Tonio



    Hence find $\displaystyle \int x tan^{-1} (x^2) dx $

    i do integration by parts with u=$\displaystyle tan^{-1} (x^2)$
    $\displaystyle \frac{du}{dx}=\frac{1}{x^4+1}$
    v=$\displaystyle \frac{x^2}{2}$

    Then i got $\displaystyle 0.5x^2 tan^{-1} (x^2) - \int\frac{x^2}{2x^4 +2}$

    i got stuck here. pls help!
    .
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