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!

Substitutions with e

  1. Sep 16, 2011 #1
    1. The problem statement, all variables and given/known data

    \begin{equation} \int \frac{e^{x}+1}{e^{x}}\end{equation}

    2. Relevant equations



    3. The attempt at a solution

    This problem comes out of the substitution section of my book, and the answer in the back of the book is [itex]x-e^{x}+C[/itex]

    I started by changing the form to this:

    \begin{equation}\int \frac{1}{e^{x}}e^{x}\end{equation}

    I set u = [itex]e^{x}[/itex] and du is the same.

    That left me with:

    \begin{equation}\int \frac{1}{u} du\end{equation}

    Integrating [itex]\frac{1}{u}[/itex] gives ln|u|

    So plugging u back in gives [itex]ln|e^{x}|[/itex]

    I plugged some test numbers into my answer and the book answer and they come out pretty close, so I thought both of them may be right, but I still can't figure out how to get [itex]x-e^{-x}[/itex].

    Any ideas?
     
  2. jcsd
  3. Sep 16, 2011 #2

    dynamicsolo

    User Avatar
    Homework Helper

    For one thing, you changed the problem when you divided:

    [tex] \frac{e^{x}+1}{e^{x}} = \frac{e^{x}}{e^{x}} + \frac{1}{e^{x}} = 1 + \frac{1}{e^{x}} [/tex]

    And I don't think you quite copied the answer from the book correctly...
     
  4. Sep 16, 2011 #3
    Ok, after reviewing it some more I found my problem. The derivative of [itex]e^{-x}[/itex] is actually [itex]-e^{-x}[/itex]. I didn't realize I had to bring down the negative sign.

    And I did copy the answer wrong, it should be a -x exponent. My bad.

    Thanks for the help.
     
  5. Sep 16, 2011 #4
    Ok, I was wrong again. I thought I had it figured out.

    I have:

    \begin{equation}\int 1 + e^{-x}\end{equation}

    I set u = [itex]e^{-x}[/itex] and du = [itex]-e^{-x}[/itex] but then there is nothing for du to replace.

    I'm lost again.
     
  6. Sep 16, 2011 #5
    In this case, you would not do a u-sub. You would split up the integral (as I say it) like so:

    [itex]\int1[/itex]dx+[itex]\int e^{-x}[/itex]dx

    You should be able to integrate it from there.
     
    Last edited: Sep 16, 2011
  7. Sep 16, 2011 #6

    Mark44

    Staff: Mentor

    You need to get into the habit of including the differential, dx in this case. When you start out with integration techniques, as you are doing here, it's not so crucial. However, as the problems get more difficult, I guarantee that omitting the differential will come back and bite your hind end.
     
  8. Sep 16, 2011 #7

    dynamicsolo

    User Avatar
    Homework Helper

    Besides the problems that can create for yourself, it is very likely that homework and exam graders will dock you on every problem where you omit it, or otherwise use notation incorrectly. (What you do in your own "scratch work" is your business, though being sloppy can cost you at some point...)
     
  9. Sep 17, 2011 #8
    Yes, I can integrate it from here, but since the problem is under the substitution review section in my book, I thought I would try to figure out how to do it with substitution.

    I usually write the dx whenever I work my problems out on my official paper, but I do avoid it on my scratch work. I guess I should break that habit.
     
  10. Sep 17, 2011 #9

    Mark44

    Staff: Mentor

    When you evaluate [itex]\int e^{-x}~dx[/itex], you will need to use a simple substitution.
    Yes, you should. When you get to trig substitutions, omitting the differential will definitely cause errors.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook