What is the antiderivative of (ln n)^2/n^2? Any help would be greatly appreciated. Thanks.
For $\displaystyle \frac{log^2(n)}{n^2}$, integrate by parts,
$\displaystyle \int f dg = f g- \int g. df$, where
$\displaystyle f = log^{2}(n) $ , $\displaystyle dg = \frac{1}{n^2} dn$,
$\displaystyle df = \frac{2 log(n)}{n} dn $, $\displaystyle g = \frac{-1}{n}$
Try doing the rest and post if you have problems.
You need to use integration by parts.
I.e. $\displaystyle \int{u\,dv} = u\,v - \int{v\,du}$.
Let $\displaystyle u = (\ln{n})^2$ so that $\displaystyle du = \frac{2\ln{n}}{n}$.
Let $\displaystyle dv = \frac{1}{n^2}$ so $\displaystyle v = -\frac{1}{n}$.
Therefore $\displaystyle \int{\frac{(\ln{n})^2}{n^2}\,dn} = -\frac{(\ln{n})^2}{n} - \int{-\frac{2\ln{n}}{n^2}\,dn}$
$\displaystyle = -\frac{(\ln{n})^2}{n} +2 \int{\frac{\ln{n}}{n^2}\,dn}$.
Now you'll need to use integration by parts again.
Let $\displaystyle u = \ln{n}$ so that $\displaystyle du = \frac{1}{n}$.
Let $\displaystyle dv = \frac{1}{n^2}$ so that $\displaystyle v = -\frac{1}{n}$.
So $\displaystyle \int{\frac{(\ln{n})^2}{n^2}\,dn} = -\frac{(\ln{n})^2}{n} +2 \int{\frac{\ln{n}}{n^2}\,dn}$
$\displaystyle = -\frac{(\ln{n})^2}{n} + 2\left(-\frac{\ln{n}}{n} - \int{-\frac{1}{n^2}\,dn}\right)$
$\displaystyle = -\frac{(\ln{n})^2}{n} - \frac{2\ln{n}}{n} + 2\int{\frac{1}{n^2}\,dn}$
$\displaystyle = -\frac{(\ln{n})^2}{n} - \frac{2\ln{n}}{n} - \frac{2}{n} + C$