Complex integral without cauchy

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SUMMARY

The discussion centers on evaluating a complex integral without employing Cauchy's theorem, which simplifies the process to yield a result of 6πi. The user expresses difficulty in finding a straightforward method, having encountered multiple variable changes that complicate the solution. A suggested substitution is provided: u = 2e^(it) + 1, which may simplify the integral's evaluation. The numerators involved are consistently of the form i(2e^(it))^n.

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Daniiel
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Hey,

I've been trying to do this integral without cauchy's theorem (with the theorem i get 6ipi in like 2 steps). I am stuck at this point, I have found afew ways to do the integral I am stuck on but they all involve multiple variable changes and I was wondering if there is a simple way to do this that I'm not seeing.

[PLAIN]http://img832.imageshack.us/img832/7958/intup.jpg sorry typo, its not supposed to be 1/2 ln(1+2e^(it)) just ln(1+2e^(it))

Thanks in advanced
 
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It helps to note that the numerators are all of the form i(2eit)n. Try using the substitution u=2eit+1 on all of them.
 

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