Antiderivative

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

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

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

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