What is intr(lnr)^2dr?

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

May 4, 2018

$\int r {\ln}^{2} r \setminus d r = \setminus \frac{{r}^{2}}{4} \left(2 {\ln}^{2} r - 2 \ln r + 1\right) + C$

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

$J : = \int {\ln}^{2} r \setminus d r$.

To solve it, we substitute $r \setminus \mapsto {e}^{x}$, which yields

${\left[\int {e}^{x} {x}^{2} \setminus d x\right]}_{x = \ln r}$.

Using integration by parts (twice), we get

${\left[{e}^{x} {x}^{2} - 2 \int {e}^{x} x \setminus d x\right]}_{x = \ln r} =$

$= {\left[{e}^{x} {x}^{2} - 2 {e}^{x} x + 2 \int {e}^{x} \setminus d x\right]}_{x = \ln r} =$

$= {\left[{e}^{x} {x}^{2} - 2 {e}^{x} x + 2 {e}^{x}\right]}_{x = \ln r} + C =$

$= r {\ln}^{2} r - 2 r \ln r + 2 r + C$.

The second preparatory integral is even simpler:

$K : = \int r \ln r \setminus d r =$

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

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

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

$I : = \int r {\ln}^{2} r \setminus d r$.

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

$r J - \int J \setminus d r =$

$= {r}^{2} {\ln}^{2} r - 2 {r}^{2} \ln r + 2 {r}^{2} - \int r {\ln}^{2} r - 2 r \ln r + 2 r \setminus d r =$

$= {r}^{2} {\ln}^{2} r - 2 {r}^{2} \ln r + 2 {r}^{2} - \int r {\ln}^{2} r \setminus d r + 2 \setminus \int r \ln r \setminus d r - 2 \int r \setminus d r =$

$= {r}^{2} {\ln}^{2} r - 2 {r}^{2} \ln r + 2 {r}^{2} - I + 2 K - 2 \int r \setminus d r =$

$= {r}^{2} {\ln}^{2} r - 2 {r}^{2} \ln r + 2 {r}^{2} - I + {r}^{2} \ln r - {r}^{2} / 2 - {r}^{2} + C =$

$= - I + {r}^{2} {\ln}^{2} r - {r}^{2} \ln r + {r}^{2} / 2 + C$.

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

$I = \setminus \frac{{r}^{2}}{4} \left(2 {\ln}^{2} r - 2 \ln r + 1\right) + C$.