Integration question help

It's probably very simple, but I just can't seem to get the right answer. If anybody could help me out it'd be much appreciated, thanks.

I'm not sure how to go about writing this, but here it goes:

Integral (limits are ln4 and 0) e^x(e^x +1)/(e^x -1) dx

And the +1 and -1 are seperate from the power.

Thanks!

Michael Collins

Not that simple at all. In fact, with the limits as given the answer is undefined. Are you sure you've written it down correctly? Should one of the exponentials have a negative exponent by any chance?

The way to go about the problem as given is with a substitution:

w = e^x

and solve.

You should get the antiderivative to be

w + 2*ln(w-1) + C.
(Just pretending it's an indefinite intergral for the moment).

The new limits (with this substitution) are 4 and 1. We can see that if we put w = 1 into the above antiderivitive, you get ln(0) which is undefined.

Michael Collins

Michael Collins wrote: »

w = e^x

and solve.

You should get the antiderivative to be

w + 2*ln(w-1) + C.
(Just pretending it's an indefinite intergral for the moment).

I think I solved it, you let w = e^x, shouldn't you have let it equal e^x -1 (or e^x +1)? then when you sub in your limits, you have to subtract (or add) 1 to it. I.e. e^ln4 -1 = 3, whereas you said 4? But then again I don't know, I wouldn't be the best at maths! Anyway I got an answer of 3, and I don't have a clue if its right or wrong!

Cokehead Mother

Either subsitution is fine. I disagree with your answer of 3. I've posted the method anyway so you can see where you went wrong and how the e^x sub works.

Anyways like Michael Collins said ln0 is undefined so you must have taken it down incorrectly or there's a mistake in the book or whatever.

Cokehead Mother

Cokehead Mother wrote: »

Either subsitution is fine. I disagree with your answer of 3. I've posted the method anyway so you can see where you went wrong and how the e^x sub works.

Anyways like Michael Collins said ln0 is undefined so there must you must have taken it down incorrectly or there's a mistake in the book or whatever.

Thanks a million for doing it out in such detail, really helped.

I see where I went wrong anyway, I used the limits 3 and 0 (used the substitution correctly). And at the 2/u part, I brought u up over the line and it became 2u^-1. I completely forgot about it actually being 2ln(u), not 2ln^-1 (When I integrated 2ln^-1, it became 0, so that is where I got 3 as my answer).

My guess is the teacher wrote it down wrong, she gave us 50 integration questions written on one sheet from past mock papers, so I presume she took down a limit wrong.

Thanks again for all your help.

Oh, just out of curiosity Cokehead_Mother, what program did you use to type out the equations? Thanks.

Cokehead Mother

TeXnicCentre.