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

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

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

    Then i got 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 \frac{d}{dx} [tan^{-1} (x^2)] I found it to be \frac{1}{1+x^4}



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

    Tonio



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

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

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

    i got stuck here. pls help!
    .
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