- #1
zak1989
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Homework Statement
Ok the problem is:
∫-1/(4x-x^2) dx
The answer in the back of the book is:
(1/4)ln(abs((x-4)/x))) + C
Homework Equations
I think this would be used somehow:
∫ du/(a^2-u^2) = 1/2a ln(abs((a+u)/(a-u))) + C
The Attempt at a Solution
∫-1/(4x-x^2) dx
∫-1/(-x^2+4x) dx
∫-1/-(x^2-4x) dx
∫1/(x^2-4x) dx
∫1/(x^2-4x+4-4) dx
∫1/((x-2)^2-4) dx
∫1/((x-2)^2-2^2) dx
At this point I'm stumped. No matter what I try I can seem to get it to look like the book answer.
If I move the bottom part around and use u-substitution I get:
∫1/(u^2-2^2) du let u=x-2 and du = dx
∫1/(-2^2+u^2) du
∫1/-(2^2-u^2) du
∫-1/(2^2-u^2) du
-∫1/(2^2-u^2) du
I will get a negative number which isn't what I want. What am I doing wrong? I would greatly appreciate any help. Thanks :)