# Antiderivative

• April 8th 2010, 09:39 PM
vlodge
Antiderivative
What is the antiderivative of (ln n)^2/n^2? Any help would be greatly appreciated. Thanks.
• April 8th 2010, 09:51 PM
harish21
Quote:

Originally Posted by vlodge
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.
• April 8th 2010, 09:58 PM
Prove It
Quote:

Originally Posted by vlodge
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 $\int{\frac{(\ln{n})^2}{n^2}\,dn} = -\frac{(\ln{n})^2}{n} +2 \int{\frac{\ln{n}}{n^2}\,dn}$

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

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

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