Question

What is `intr(lnr)^2dr`?

Should you use u-substitution and then integration by parts? This method isn't working for me.

Answer 1

`int r ln^2 r\ d r = \frac{r^2}{4} ( 2 ln^2 r - 2 ln r + 1 ) + C`

Explanation:

Before we have a look at the integral that we are actually interested in, let's evaluate two simpler ones that will show up in the course of our calculations.

The first of them is

`J := int ln^2 r\ d r`.

To solve it, we substitute `r \mapsto e^x`, which yields

`[ int e^x x^2\ d x ]_{x = ln r}`.

Using integration by parts (twice), we get

`[ e^x x^2 - 2 int e^x x\ d x ]_{x = ln r} =`

`= [ e^x x^2 - 2 e^x x + 2 int e^x\ d x ]_{x = ln r} =`

`= [ e^x x^2 - 2 e^x x + 2 e^x ]_{x = ln r} + C =`

`= r ln^2 r - 2 r ln r + 2 r + C`.

The second preparatory integral is even simpler:

`K := int r ln r\ d r =`

`= \frac{r^2 ln r}{2} - \frac{1}{2}int r \ d r =`

`= \frac{r^2 ln r}{2} - \frac{r^2}{4} + C`.

Now we are ready to tackle the integral that we actually care about:

`I := int r ln^2 r\ d r`.

Using integration by parts, where `u' = ln^2 r` and `v = r`, we see that this is the same as

`r J - int J\ d r =`

`= r^2 ln^2 r - 2 r^2 ln r + 2 r^2 - int r ln^2 r - 2 r ln r + 2 r\ d r =`

`= r^2 ln^2 r - 2 r^2 ln r + 2 r^2 - int r ln^2 r \ d r + 2 \int r ln r\ d r - 2 int r\ d r =`

`= r^2 ln^2 r - 2 r^2 ln r + 2 r^2 - I + 2 K - 2 int r\ d r =`

`= r^2 ln^2 r - 2 r^2 ln r + 2 r^2 - I + r^2 ln r - r^2/2 - r^2 + C=`

`= -I + r^2 ln^2 r - r^2 ln r + r^2 / 2 +C`.

However, this means that we have derived the equation

`I = -I + r^2 ln^2 r - r^2 ln r + r^2 / 2 +C`,

which is easily solved with

`I = \frac{r^2}{4} ( 2 ln^2 r - 2 ln r + 1 ) + C`.