Question

How do you integrate `int6x ln x dx` from 2 to 4?

Answer 1

`int_2^4 \ 6xlnx \ dx = 84ln2-18 ~~ 40.224 \ (3 \ dp)`

Explanation:

We seek:

`I = int_2^4 \ 6xlnx \ dx`

We can then apply Integration By Parts:

Let # { (u,=lnx, => (du)/dx,=1/x), ((dv)/dx,=6x, => v,=3x^2 ) :}#

Then plugging into the IBP formula:

`int \ (u)((dv)/dx) \ dx = (u)(v) - int \ (v)((du)/dx) \ dx`

We have:

`int _2^4 (lnx)(6x) \ dx = [(lnx)(3x^2)]_2^4 - int_2^4 \ (3x^2)(1/x) \ dx`

`:. I = [3x^2lnx]_2^4 - int_2^4 \ 3x \ dx`

`\ \ \ \ \ \ \ = 3{16ln4-4ln2} - [(3x^2)/2]_2^4`

`\ \ \ \ \ \ \ = 3{16ln2^2-4ln2} -3/2 [x^2]_2^4`

`\ \ \ \ \ \ \ = 3{32ln2-4ln2} -3/2 {16-4}`

`\ \ \ \ \ \ \ = 3{28ln2} -3/2 * 12`

`\ \ \ \ \ \ \ = 84ln2 - 18`