Question

How do you integrate `[((3x^2)+2x)/x^2]dx` from 4 to 2?

Answer 1

Here's my version:
(no guarantee that it's right)

`int_4^2(3x^2+2x)/(x^2) dx`

`=int_4^2 ( (3x^2)/(x^2) + (2x)/(x^2))dx`

`= int_4^2 (3x^2)/(x^2)dx + int_4^2 (2x)/(x^2)dx`

`= 3int_4^2(1)dx + 2int_4^2(1/x) dx`

`= 3*( x|_4^2 ) + 2*(ln|x||_4^2)`

`=3( 2-4) + 2 (ln(2)-ln(4))`

`=-6 + 2(0.693147 - 1.386794)`

`=-6 - 1.3863`

`= -7.3863`