I[(e^x+1)/(e^x-1)dx]

[courteous]

\(\displaystyle \int \frac{e^x+1}{e^x-1}dx\)

\(\displaystyle =\int \frac{e^x-1+2}{e^x-1}dx=\int 1+\frac{2}{e^x-1}dx=x+\frac{2}{e^x}\log|e^x+1|+C\)

Official solution: \(\displaystyle 2\log|e^x+1|-x+C\)

I've checked it all over, and I must be missing something rather obvious. Why is there \(\displaystyle -x\), plus, why isn't there \(\displaystyle \frac{1}{e^x}\) added at log (when you differentiate \(\displaystyle \log|e^x+1|\) don't you get \(\displaystyle \frac{1}{e^x+1}e^x\))?

 

[Deleted member 4993]

\(\displaystyle \int \frac{e^x+1}{e^x-1}dx\)

\(\displaystyle =\int \frac{e^x-1+2}{e^x-1}dx=\int 1+\frac{2}{e^x-1}dx=x+\frac{2}{e^x}\log|e^x+1|+C\) <<< How did it become |e^x+1| in the denominator?
Official solution: \(\displaystyle 2\log|e^x+1|-x+C\)

I've checked it all over, and I must be missing something rather obvious. Why is there \(\displaystyle -x\), plus, why isn't there \(\displaystyle \frac{1}{e^x}\) added at log (when you differentiate \(\displaystyle \log|e^x+1|\) don't you get \(\displaystyle \frac{1}{e^x+1}e^x\))?

I think your problem was:

"\(\displaystyle \int \frac{e^x-1}{e^x+1}dx\)

instead of what you posted - at least your book solves that problem.

 

[courteous]

You have almost convinced me, so I went again to check... no, I've jotted it down correctly, and besides, Mathematica solves it the same way (as the textbook, not me, of course )!

In[1]: Integrate[(E^x + 1)/(E^x - 1), x]
Out[1]: -x + 2 Log[-1 + e^x]

 

[Deleted member 4993]

Look at mathematica solution again - it is different from what you wrote as "official" solution (it has -1 as opposed +1)

Your solution is incorrect because - if you differentiate "your" answer - you do not get the original function back.

\(\displaystyle \int\frac{e^x+1}{e^x-1} dx\)

\(\displaystyle =\, \int\frac{2e^x - (e^x-1)}{e^x-1} dx\)

\(\displaystyle =\, \int\frac{2e^x}{e^x-1} dx \, - \, \int dx\)

\(\displaystyle = \, 2ln(|e^x-1)| \, - \, x \, + \, C\)

 

[courteous]

\(\displaystyle \int \frac{e^x+1}{e^x-1}dx\)

\(\displaystyle =\int \frac{e^x-1+2}{e^x-1}dx=\int 1+\frac{2}{e^x-1}dx=x+\frac{2}{e^x}\log|e^x+1|+C\) <<< How did it become |e^x+1| in the denominator?
Official solution: \(\displaystyle 2\log|e^x+1|-x+C\)

I've checked it all over, and I must be missing something rather obvious. Why is there \(\displaystyle -x\), plus, why isn't there \(\displaystyle \frac{1}{e^x}\) added at log (when you differentiate \(\displaystyle \log|e^x+1|\) don't you get \(\displaystyle \frac{1}{e^x+1}e^x\))?

I've mistyped the +. I've actually solved it as \(\displaystyle x+\frac{2}{e^x}\log|e^x-1|+C\)

Actual official solution (corrected +): \(\displaystyle 2\log|e^x-1|-x+C\) (Hope I didn't make another plus/minus error :wink: .)

Anyway, how is my derivation (that \(\displaystyle \frac{e^x+1}{e^x-1}=\frac{e^x-1+2}{e^x-1}\)) different from the one you suggested (which is of course correct)?

 

[Deleted member 4993]

Anyway, how is my derivation (that \(\displaystyle \frac{e^x+1}{e^x-1}=\frac{e^x-1+2}{e^x-1}\)) different from the one you suggested (which is of course correct)?

I am sure you can see how it is different!

Your process of reduction is not wrong - only that you cannot integrate it, after you have done it your way.

d/dx [1/e^x * ln(e^x -1] = ????

 

[courteous]

I do not understand . \(\displaystyle =\int \frac{e^x-1+2}{e^x-1}dx=\int dx+\int\frac{2}{e^x-1}dx=...\) Isn't the last fraction integrable?

d/dx [1/e^x * ln(e^x -1)] = ????

= -e^(-x) * ln(e^x -1) + 1/(e^x-1)
[If you meant "differentiate" with d/dx.]

Ahhh... now I get it: you can't just put in one extra \(\displaystyle \frac{1}{e^x}\) to "make it right", as the derivative of the whole must lead back to what is under the integral. Thank you!