Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook