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

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Answer 1 · 2015-05-06T17:47:11

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#