Question

Integrate the attached problem?

Answer 1

Can you do a U substitution, try U=exp(2x)-1

Answer 2

\[\Large\rm \int\limits \frac{1}{\sqrt{(e^x)^2-1}}dx\]If you write it like this^
It might be a little easier to work with.

Answer 3

or actually u=root (e^2x-1)

Answer 4

u-sub into trig sub would work nicely I bet.
u=e^x

Answer 5

then du = e^x . What do I do next ?

Answer 6

and I'm working on paper with the substitution that xapproach suggested...

Answer 7

\[\Large\rm du=e^x dx\qquad\to\qquad \frac{du}{e^x}=dx\]\[\Large\rm \frac{du}{u}=dx\]

Answer 8

Gives you something like this, yes?
\[\Large\rm \int\limits \frac{1}{\sqrt{(u)^2-1}}\left(\frac{du}{u}\right)\]

Answer 9

Try this,u^2=e^2x-1, then udu=e^2xdx

Answer 10

ah, that would be arcsec u if the u in the bottom is an absolute value.. Is that correct? But is it absolute value?

Answer 11

Woah woah simmer down.
See the u multiplying the sqrtroot in front?
So it won't necessarily give us arcsecant nice and clean like that.

Answer 12

\[\Large\rm \int\limits\limits \frac{1}{\sqrt{(u)^2-1}}\left(\frac{du}{u}\right)=\int\limits\limits \frac{1}{u\sqrt{(u)^2-1}}~du\]

Answer 13

Oh you need the x out front for the arcsecant haha.
Sorry I forgot some of those weird integrals :)

Answer 14

u is positive as e^2x cannot be negative. So, arcsecant can be the integral.

Answer 15

@zepdrix this is the integral, I'm saying if it's abs value

Answer 16

oh ok @zepdrix xDD yep I had some hard time solving this!

Answer 17

still can't solve it :P

Answer 18

The absolute thing on the x?
Ehh I dunno >.<

Ya arcsecant sounds good though.
Did the u-sub make sense though? :o

Answer 19

Yep it did make sense! Maybe it's arcsecant... Well I'll just leave it blank and skip then?

Answer 20

looks @xapproachesinfinity 's first substitution gives nice simplification...

Answer 21

it substitutes the entire thing inside radical so that we don't have to bother about second substitutions...

Answer 22

@ganeshie8 yep I'm trying as well. Thsi one right?
-Try this,u^2=e^2x-1, then udu=e^2xdx

Answer 23

u guys when you did the substitution you diid for expx not exp2x

Answer 24

Yah we rewrote \(\Large\rm e^{2x}\) as \(\Large\rm (e^x)^2\) and made a substitution from there :)
Just a different approach.

Answer 25

Oh I see

Answer 26

it looks an arctan, no?

Answer 27

Yes, it is \(\tan^{-1} \sqrt{e^{2x} - 1}\) with substitution \(u = \sqrt{e^{2x} - 1}\)

Answer 28

Oh you're right about the u @Shailkumar and @xapproachesinfinity. It does make it a litle easier!

Answer 29

So I got arctan (e^(2x) -1) + C. That's the answer right? :D

Answer 30

Yes

Answer 31

thank you so much for helping me!
@ganeshie8
@xapproachesinfinity
@zepdrix
@ShailKumar

:)) Thank you <3

Answer 32

yeah, that what I got