1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Find the antiderivative

  1. Dec 14, 2008 #1
    1. The problem statement, all variables and given/known data
    Find the antiderivative of each derivative:

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

    3. The attempt at a solution

    I can't entirely figure out how to get myself started here.
    I tired splitting up the derivative so that it's

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

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

    I think I should try substitution but I can't figure out what value to make u.

    If anyone could help me, that would be fantastic. I don't think I'm going quite in the right direction.
  2. jcsd
  3. Dec 14, 2008 #2


    User Avatar
    Science Advisor
    Homework Helper

    This would be a good time to learn about hyperbolic function like sinh(x) and cosh(x), but you can also do it directly. Take u=e^x. So e^(-x)=1/u.
  4. Dec 14, 2008 #3
    I would suggest thinking about sinh and cosh... and then relying on some trigonometric formulas. I am not good with these hyperbolic things.

    edit: oops.. dick already mentioned it while I was solving using those sinhs...
  5. Dec 14, 2008 #4
    Thanks. Technically, for this exam I'm studying for, we're not supposed to know those hyperbolic functions- so I've got to make use of substitution.

    Okay, so:

    I split the derivative into the two parts:(e^x)[(e^x)-(e^-x)]^-2 and (e^-x)[(e^x)-(e^-x)]^-2 and made u=e^x so du=(e^x)dx and e^-x=1/e^x

    So the antiderivative of (e^x)[(e^x)-(e^-x)]^-2 = [-1/u]-[(u^3)/3]

    so now I'm looking for the antiderivative of (e^-x)[(e^x)-(e^-x)]^-2

    So I end up with [(4(u^3))/3]-4ln|u|

    Which makes the antiderivative=
    then I change the u's back.


    so I get [-1/(e^x)]-[(e^3x)/3]+[(4(e^3x))/3]-4x

    it gives me: [-e^-x]-[e^3x]-x

    ...and the book tells me I'm still wrong.

    Can anyone help me out??
  6. Dec 14, 2008 #5


    User Avatar
    Science Advisor
    Homework Helper

    You are already way off track. I don't see how you got that antiderivative. Just do the substitution without splitting anything up. Show the whole integral in terms of u before you go leaping forward.
  7. Dec 14, 2008 #6


    User Avatar
    Science Advisor
    Homework Helper

    Stop there. (u-1/u)^(-2) is not equal to (u^(-2)-u^(2)). You can't distribute powers like that. You'll never get to the end correctly with algebra like that. And dx is not equal to du. du=e^(x)*dx=u*dx.
  8. Dec 14, 2008 #7


    User Avatar
    Science Advisor
    Homework Helper

    I told you. You are doing bad algebra. (u-1/u)^(-2) is not equal to (u^(-2)-u^(2)). If you don't know what to do with it, don't do anything with it.
    Last edited: Dec 14, 2008
  9. Dec 14, 2008 #8


    User Avatar
    Science Advisor
    Homework Helper

    I'll give you a hint. (u+1/u)=(u^2+1)/u. (u-1/u)^2=((u^2-1)/u)^2=(u^2-1)^2/u^2. I'm trying to write these expressions as polynomials in u so I can factor them and use partial fractions.
  10. Dec 14, 2008 #9
    Thanks Dick, you're really helping me right now. I have a lot of trouble with math and so I'm getting very turned around on this problem.

    That said, I don't really understand your hint. I'm just getting really frustrated and confused with this.

    I'm going to try to go through it step for step. Maybe you (or anyone) can point out to me
    where that hint could help me?

    I'm looking for the antiderivative of [(e^x)+(e^-x)]/[(e^x)-(e^-x)]^2

    and so u=e^x
    du=(e^x)dx ; du=udx
    e^-x = 1/e^x

    okay... I= [[(u)+(1/u)]/[(u)-(1/u)]^2]dx
    because there's a u in the numerator can I combine that with the dx to make the du?

    I= [(u^-1)/[u-(u^-1)]^2] du

    Okay- so I'm going to look at the denominator for a second:
    I have to factor that out, right? to (u^2)-2-(1/u^2)

    I= (1/u)/[(u^2)-2-(1/u^2)] du

    can I multiply [(1/u^2)-(1/2)-u^2)] by (1/u) ?
    is this closer to the right track???

    I'm really sorry I sound so inept, I just don't understand and I am so confused!

    Any help would be a life-saver. Thanks.
  11. Dec 14, 2008 #10


    User Avatar
    Science Advisor

    No, you certainly can't do that- the "u" you are using is added, not multiplied What you could do that might help is multiply both numerator and denominator by u2.

  12. Dec 14, 2008 #11


    User Avatar
    Science Advisor
    Homework Helper

    What's making you sound inept is obvious algebraic mistakes. Don't make them. I'll give you a target to shoot for. The integrand becomes (u^2+1)*du/(u^2-1)^2 if you play your cards right. That was supposed to be the easy part. Now you have to use partial fractions on that. Which also needs some algebra skills without making mistakes on every line. Try the first one first, ok?
  13. Dec 15, 2008 #12


    User Avatar
    Science Advisor
    Homework Helper

    Try the substitution

    [tex] u=e^{x}-e^{-x} [/tex].

    What is "du" equal to ??
  14. Dec 15, 2008 #13


    User Avatar
    Science Advisor
    Homework Helper

    Good idea. Much less work.
  15. Dec 15, 2008 #14
    Ah thank you. I finally managed to figure it out.

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

    u=e^x - e^-x
    du= e^x - e^-x (-1) dx
    du= - e^x + e^-x dx

    I= 1/u^2 (du)
    I= u^-2 (du)
    I= -1 u^-1
    I= 1/-u + c

    p= -1/( e^x - e^-x) + C

    Thanks for all the suggestions and help- I definitely just hit a ridiculous mental block with this paticular question. Thanks again!
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook