Question

Need help integrating 1/(e^2x) -1. Can anyone help?

Answer 1

Is this your question
\[ \int \frac{1}{e^{2x}-1} dx\]

Answer 2

Substitute u=e^2x
Then it is easy

Answer 3

alsunshine will you be able to solve it?

Answer 4

no the dx isn't on top I'm not sure what to do

Answer 5

dx will come automatically
\[ \int \frac{1}{e^{2x}-1} dx\]

Answer 6

i don't think so

Answer 7

as we are integrating with respect to x

Answer 8

i mean the du will equal 2e^2xdx. I'm not sure what to do with that.

Answer 9

u= e^2x ,
so
du=2udx
can you solve it now?

Answer 10

i think you have work to do. are going to have two rewrite either as
\[\int\frac{1}{(e^x+1)(e^x-1)}dx\] or as
\[\frac{1}{2}\int \frac{1}{e^u-1}\] and then use partial fractions

Answer 11

@ash you may have something, but i don't see it. can you explain?

Answer 12

We have
\[ u=e^{2x} \]
\[ du =2 e^{2x} dx\]
but u=e^2x
so
\[ du = 2u dx\]
Now
the integral is
\[ \int \frac{ 2u }{u-1} du\]

this can be written as
\[ \int \frac{ 2(u-1+1) }{u-1} du\]
so we get
\[ \int 1+\frac{ 2(1) }{u-1} du\]
we get
\[u + 2\ln(u-1)+c\]
where u=e^2x

Answer 13

sunshine you got it?

Answer 14

looks good to me!

Answer 15

thanks so much guys. that looks like a mess

Answer 16

didn't you understand?

Answer 17

understood until you rewrote as 2(u-1+1)/u-1?

Answer 18

how did you get that from 2u

Answer 19

Sorry I was outside
we have
\[2(u)\]
subtract and add 1 to u
\[2(u-1+1)\]
This doesn't change any thing

Answer 20

now
\[\frac{2(u-1+1)}{u-1}=\frac{2(u-1)}{u-1}+\frac{2(1)}{u-1}=> 2+ \frac{2}{u-1}\]

Answer 21

oh ok! Thanks so much for your help