1. Not finding help here? Sign up for a free 30min 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!

Integration Help

  1. Jan 19, 2009 #1
    1. The problem statement, all variables and given/known data
    [itex]
    \int {\frac {1}{\sqrt {{e^{x}}+1}}} dx
    [/itex]


    3. The attempt at a solution

    I tried integration by parts, using [itex]{\frac {1}{\sqrt {{e^{x}}+1}}}[/itex] as u and dx as dv. This seemed to make the problem more complicated. Using e^x+1 as a u substitution won't work either. Any help getting on the right path would be greatly appreciated!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 19, 2009 #2

    rock.freak667

    User Avatar
    Homework Helper

    You could try hyperbolic trig sub. such as ex=sinhu
     
  4. Jan 19, 2009 #3
    I don't know what hyperbolic trig sub. is yet. We've just gotten to integration by parts.
     
  5. Jan 19, 2009 #4

    rock.freak667

    User Avatar
    Homework Helper

    Have you done normal trig substitutions? As ex=tan2u looks like it would work out fine
     
  6. Jan 19, 2009 #5
    Nope. Maybe it's a problem I'm just supposed to try and not succeed at.
     
  7. Jan 19, 2009 #6

    rock.freak667

    User Avatar
    Homework Helper

    Did you learn partial fractions yet? If you did try putting [itex]u^2=e^x+1[/itex] and post back what you tried.
     
  8. Jan 19, 2009 #7
    I'm not sure of the procedure when using that technique. if u^2=e^x+1, then does the integral become 1/u? Then the answer would just be ln(u)... I dont think thats right
     
  9. Jan 19, 2009 #8

    rock.freak667

    User Avatar
    Homework Helper

    u^2=e^x +1
    then 2u du =e^x dx and e^x=?
    and so dx =?
     
  10. Jan 21, 2009 #9
    This problem is driving me nuts. There was a typo earlier, it's actually e^x-1.

    Expanding on what you said, if u^2=e^x-1
    2udu=e^xdx and dx=(2udu/e^x). Moving all of this stuff around is just fine and dandy but I don't know where to put it back in.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Integration Help
  1. Integral help (Replies: 4)

  2. Integration Help (Replies: 5)

  3. Integration help! (Replies: 3)

  4. Help with an integral (Replies: 1)

  5. Integrals help (Replies: 7)

Loading...