What is intr(lnr)^2dr?

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

1 Answer
May 4, 2018

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.