what is the integral of (1+e^x)/(1-e^x) dx?

Question

please explain. thanks a bunch

Answer 1

Use long division again.
(1+e^x)/(1-e^x) = 1 + (2e^x)/(1-e^x)
The integral of 1 is x
To integrate 2e^x/(1-e^x) dx , use substitution. Let u = 1 - e^x, then du = -e^x dx or -du = e^x dx
2e^x/(1-e^x) dx becomes
-2/u du
= -2 ln |u|
= -2 ln |1-e^x|
Answer: x - 2 ln|1-e^x| + C

Answer 2

x - (e^2x)/2

Where 'S' is an integral symbol:

(1+e^x)(1-e^x)dx = 1 - e^2x

From Standard Mathematical tables 27th edition:
p236, formula 12:

S e^(ax) dx = (e^(ax))/a

Hence applied to the above where a = 2:

S 1 - e^(2x) dx = S 1 dx - S -e^(2x) dx = x - (e^2x)/2

This is the correct answer. Don't give a thumbs down unless you can disprove it. Mathgirl is incorrect on this one.

Well, I got 2 thumbs down and there are 2 INCORRECT answers below me. I suppose you bozos think you are being incognito. Think again, and get your math right.

DutchProf,
It's Dr Evil, not Mr Evil,
Anyways, I'll give you the + C, since the derivative of a constant is zero; however, that constant could have originally been zero, hence + C would not be necessary. So, that is somewhat of a grey area. But the + C is common.

However, I do not follow your logic about involving division, etc. The asker asked for the integral. That is what I gave him. There is no desire by the asker to go into hyperbolic cotangents.

Answer 3

I multiply numerator and denominator with e^(-x/2) and find

-[e^(x/2) + e^(-x/2)] / [e^(x/2) - e^(-x/2)] dx

which is equal to a hyperbolic cotangent,

-coth (x/2) dx = -cosh (x/2) / sinh (x/2) dx

Writing u = sinh (x/2), we have du = 1/2 cosh (x/2) dx, so the integral is

INT -2 du / u = -2 ln |u| + C

which, after substitution, becomes

-2 ln |sinh (x/2)| + C =
... = -2 ln |e^(x/2) - e^(-x/2)| + C

This is a nice symmetric expression, but if you don't care about that, you can simplify:

... = -2 ln |e^(-x/2) * {e^x - 1}| + C
... = -2 ln |e^(-x/2)| - 2 ln |e^x - 1| + C
... = x - 2 ln |e^x - 1| + C

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Btw, mr. Evil, you are right that the integral of (1 + e^x) (1 - e^x) dx, or equivalently, [1 - e^(2x)] dx, is equal to x - 1/2 e^(2x) + C. But that was not the question: the original problem involves division, not multiplication. And that is quite different.

Answer 4

this could be written as
(1-e^x + 2e^x)/ 1-e^x
then
1 + 2e^x/1-e^x
then
1 - 2(e^x/e^x -1)

split it into two parts to integrate
1 becomes simply x +c

-2(e^x/e^x -1) becomes -2(ln le^x-1l ) + c

thus the answer is x - 2ln le^x-1l + c
or... x - ln(e^x-1)^2 + c
or... x - ln(e^2x - 2e^x +1) +c