- #1
Painguy
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Homework Statement
evaluate the integral 1/(u^2 -36) from 0 to 6
does the integral converge?
Homework Equations
The Attempt at a Solution
integral 1/(u^2 -36)
integral 1/((u-6)(u+6))
Partial fraction decomposition
1/((u-6)(u+6)) = A/(u-6) + B/(u+6)
1=A(u+6) + B(u-6)
1=(A+B)u +(6A-6B)
A+B=0
A=-B
6A -6B=1
-12B=1
B=-1/12
A=1/12
1/12 int 1/(y-6) - 1/12 int 1/(y+6)
1/12 ln|y-6| - 1/12 ln|y+6|
I'm being told however by wolframalpha that it should be 1/12 ln|6-y| - 1/12 ln|y+6|
How did that happen?
I also wanted to try to use the comparison theorem to see if it converges or not.
I use the function g(x)=1/y^2 and i know that does not converge from 0 to 6, and since it is larger than f(x) then f(x) also does not converge. Did i go about that correctly?
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