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# partial fractions watch

1. find the value of the integral of 18/x^2(x+3) between the limits of 2 and 1

18/x^2(x+3) = a/x + b/x^2 + c/x+3

18 = ax(x+3) +b(x+3) +cx^2

sub in x=0 to get b = 6
x=-3 gets c = 2
x^2 gets 0 = c+a --> 0=2+a A= -2

so 18/x^2(x+3) = 6/x^2 - 2/x +2/x+3

integrating gets 2ln(x+3) - 6/x - 2lnx

sub in limits of 2 and 1

(2ln(5) - 3 -2ln(2)) - (2ln(4) - 6 -2ln(1))

2ln(5) -2ln(2) -2ln(4) + 3

using laws of logs i got 2ln5/8 + 3

but the answers say it is 3 - 2ln8/5

where have i gone wrong?
2. Oh this happened to one of my answers earlier today.

2ln5/8 is the same as - 2ln8/5 so you got it right
3. thank you - I thought it was something to do with taking the logs to the power -1 but wasnt completely sure

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