Integral of (1+e^x)/(1-e^x) dx: Simplifying and Using Substitution Method

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Homework Statement



integral of (1+e^x)/(1-e^x) dx

Homework Equations





The Attempt at a Solution


The TA said to make u = e^x
So, du = e^x dx. dx = du/e^x.
Since e^x = u

The integral now is (1+u)/(1-u)u
I am confused as to what to do after distribute the u in the bottom.
 
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Use partial fractions.
 
alright so i got A = 1 and B = 2 so now i have the integral of 1/u + 2/(1-u). Which i end up getting ln |u| - 2ln|1-u| + c. Now replacing the u for e^x i get ln e^x - 2 ln(1-e^X). Which is x-2 ln (1-e^x) + c. Is that the right answer?
 
Here's an alternate method you can use to solve this problem.

Whenever I see combinations of 1\pm e^x, I look at what happens if I pull out a factor of e^{x/2} to restore symmetry to the quantities. In this case, the integrand becomes

\frac{1+e^x}{1-e^x} = \frac{e^{x/2}(e^{-x/2}+e^{x/2})}{e^{x/2}(e^{-x/2}-e^{x/2})} = \frac{e^{-x/2}+e^{x/2}}{e^{-x/2}-e^{x/2}}

You might notice that the top and bottom can be written in terms of hyperbolic trig functions or that they are derivatives of each other to within a constant factor. In either case, with the appropriate substitution, integrating is straightforward.
 
Check it yourself and see if it's the right answer. If the derivative of your answer equals the integrand, then your answer is correct.
 
akbar786 said:
alright so i got A = 1 and B = 2 so now i have the integral of 1/u + 2/(1-u). Which i end up getting ln |u| - 2ln|1-u| + c. Now replacing the u for e^x i get ln e^x - 2 ln(1-e^X). Which is x-2 ln (1-e^x) + c. Is that the right answer?
It's easy enough to check. Try differentiating your answer and see if you recover what you started with.
 
wow that makes is so much simpler..thanks so much!
 
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