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Math Help - Antiderivative

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    Antiderivative

    What is the antiderivative of (ln n)^2/n^2? Any help would be greatly appreciated. Thanks.
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    Quote Originally Posted by vlodge View Post
    What is the antiderivative of (ln n)^2/n^2? Any help would be greatly appreciated. Thanks.
    For \frac{log^2(n)}{n^2}, integrate by parts,

     \int f dg = f g- \int g. df, where

     f = log^{2}(n) , dg = \frac{1}{n^2} dn,

    df = \frac{2 log(n)}{n} dn , g = \frac{-1}{n}

    Try doing the rest and post if you have problems.
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    Quote Originally Posted by vlodge View Post
    What is the antiderivative of (ln n)^2/n^2? Any help would be greatly appreciated. Thanks.
    You need to use integration by parts.

    I.e. \int{u\,dv} = u\,v - \int{v\,du}.


    Let u = (\ln{n})^2 so that du = \frac{2\ln{n}}{n}.

    Let dv = \frac{1}{n^2} so v = -\frac{1}{n}.


    Therefore \int{\frac{(\ln{n})^2}{n^2}\,dn} = -\frac{(\ln{n})^2}{n} - \int{-\frac{2\ln{n}}{n^2}\,dn}

     = -\frac{(\ln{n})^2}{n} +2 \int{\frac{\ln{n}}{n^2}\,dn}.


    Now you'll need to use integration by parts again.

    Let u = \ln{n} so that du = \frac{1}{n}.

    Let dv = \frac{1}{n^2} so that v = -\frac{1}{n}.


    So \int{\frac{(\ln{n})^2}{n^2}\,dn} = -\frac{(\ln{n})^2}{n} +2  \int{\frac{\ln{n}}{n^2}\,dn}

     = -\frac{(\ln{n})^2}{n} + 2\left(-\frac{\ln{n}}{n} - \int{-\frac{1}{n^2}\,dn}\right)

     = -\frac{(\ln{n})^2}{n} - \frac{2\ln{n}}{n} + 2\int{\frac{1}{n^2}\,dn}

     = -\frac{(\ln{n})^2}{n} - \frac{2\ln{n}}{n} - \frac{2}{n} + C
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