(New page: Evaluate the integral: <math> \int \frac{dx}{1+e^x} </math> I tried setting 'u' equal to '1+e^x' and 'du' equal to 'e^x dx'. Anyone know where to go from here? I'm sure there's some ki...)
 
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I tried setting 'u' equal to '1+e^x' and 'du' equal to 'e^x dx'.  Anyone know where to go from here? I'm sure there's some kind of manipulation that needs to be pulled off.  I'll tell you if I figure it out.[[User:Gbrizend|Gbrizend]]
 
I tried setting 'u' equal to '1+e^x' and 'du' equal to 'e^x dx'.  Anyone know where to go from here? I'm sure there's some kind of manipulation that needs to be pulled off.  I'll tell you if I figure it out.[[User:Gbrizend|Gbrizend]]
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Alright, I factored out a <math> e^x </math>. I set 'u' equal to <math> 1+\frac{1}{e^x} </math> and 'du' to <math> -1\frac{1}{e^x} dx </math>.  The problem looks like this: <math> \int \frac{dx}{e^x(1+\frac{1}{e^x})} </math>.  Substitute with u and du.  I got this answer: <math> -Ln|1+\frac{1}{e^x}| + C </math>

Revision as of 10:56, 29 September 2008

Evaluate the integral: $ \int \frac{dx}{1+e^x} $


I tried setting 'u' equal to '1+e^x' and 'du' equal to 'e^x dx'. Anyone know where to go from here? I'm sure there's some kind of manipulation that needs to be pulled off. I'll tell you if I figure it out.Gbrizend

Alright, I factored out a $ e^x $. I set 'u' equal to $ 1+\frac{1}{e^x} $ and 'du' to $ -1\frac{1}{e^x} dx $. The problem looks like this: $ \int \frac{dx}{e^x(1+\frac{1}{e^x})} $. Substitute with u and du. I got this answer: $ -Ln|1+\frac{1}{e^x}| + C $

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