Results 1 to 10 of 10

Math Help - grrrrr another EZ integral

  1. #1
    Junior Member
    Joined
    Jan 2009
    Posts
    52

    grrrrr another EZ integral

    <br />
\int{\frac{e^{-x}}{1 + e^{-(2x)}}}<br />

    I tried my very best to solve this but I am stuck. I really dislike my text book at all and my professor is very hard for me to understand. Does anyone have a good resource online for this type of problem? I understand all the new concepts we are supposed to be learning but I am stuck on the stuff I was supposed to learn last semester e.g. 5 years ago for me.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Oct 2008
    Posts
    77
    Thanks
    1
    Split it into two different integrals..

    \int {e^{-x}} + \int{e^{-x}/e^{-2x}}

    EDIT: LaTeX fixed.
    Last edited by Chris L T521; March 7th 2009 at 02:01 PM. Reason: fixed LaTeX
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by TYTY View Post
    <br />
\int{\frac{e^{-x}}{1 + e^{-(2x)}}}<br />

    I tried my very best to solve this but I am stuck. I really dislike my text book at all and my professor is very hard for me to understand. Does anyone have a good resource online for this type of problem? I understand all the new concepts we are supposed to be learning but I am stuck on the stuff I was supposed to learn last semester e.g. 5 years ago for me.
    Note that \int\frac{e^{-x}}{1+e^{-2x}}\,dx=\int\frac{e^{-x}}{1+\left(e^{-x}\right)^2}\,dx

    I leave it for you to show that by the substitution u=e^{-x}, you have \int\frac{e^{-x}}{1+\left(e^{-x}\right)^2}\,dx\xrightarrow{u=e^{-x}}{}-\int\frac{\,du}{1+u^2}

    Can you try the problem now?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by Scopur View Post
    Split it into two different integrals..

    \int {e^{-x}} + \int{e^{-x}/e^{-2x}}
    You can't do this. You can't split the denominator up like that!!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Oct 2008
    Posts
    77
    Thanks
    1
    Oh my bad.. i am not thinking well today apparently.. :S
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member Mentia's Avatar
    Joined
    Dec 2008
    From
    Bellingham, WA
    Posts
    134
    use the substitution: u = e^(-x)
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Jan 2009
    Posts
    52
    Quote Originally Posted by Chris L T521 View Post
    Note that \int\frac{e^{-x}}{1+e^{-2x}}\,dx=\int\frac{e^{-x}}{1+\left(e^{-x}\right)^2}\,dx

    I leave it for you to show that by the substitution u=e^{-x}, you have \int\frac{e^{-x}}{1+\left(e^{-x}\right)^2}\,dx\xrightarrow{u=e^{-x}}{}-\int\frac{\,du}{1+u^2}

    Can you try the problem now?
    your answer made this problem trivial - even to me haha. Weird thing is when I attempted to verify the derivative of e^(-x), I used google's first search result Online Derivative Calculator

    Which says that
    Function e^(-x)

    f'x = -log(e)/e^x

    pretty weird...
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member Mentia's Avatar
    Joined
    Dec 2008
    From
    Bellingham, WA
    Posts
    134
    but what is the natural log of e? 1. So it gave you the right answer. It just didnt assume the "e" you entered is actually e but just some variable.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    Jan 2009
    Posts
    52
    Quote Originally Posted by Mentia View Post
    but what is the natural log of e? 1. So it gave you the right answer. It just didnt assume the "e" you entered is actually e but just some variable.
    hm that makes sense. Given my level of frustration, I am getting nowhere fast with all this
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Member Mentia's Avatar
    Joined
    Dec 2008
    From
    Bellingham, WA
    Posts
    134
    Let me explain that one a little better.

    We know:

     \frac{d }{dx } e^{x} = e^{x}

    But what about, say, \frac{d }{dx } 2^{x}?

    Well, we can say,

    2^{x} = (e^{ln(2)})^{x} = e^{xln(2)}

    Then

    \frac{d }{dx } 2^{x} = \frac{d }{dx }e^{xln(2)}

    So then we use the chain rule:

    \frac{d }{dx }e^{xln(2)} = ln(2)e^{xln(2)} = ln(2)2^{x}

    So if we had some other constant, a, instead of 2:

    \frac{d }{dx }e^{xln(a)} = ln(a)e^{xln(a)} = ln(a)a^{x}

    So now you can see that this is the general form of a derivative of a^{x}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: August 31st 2010, 07:38 AM
  2. Replies: 1
    Last Post: June 2nd 2010, 02:25 AM
  3. Replies: 0
    Last Post: May 9th 2010, 01:52 PM
  4. Replies: 0
    Last Post: September 10th 2008, 07:53 PM
  5. Replies: 6
    Last Post: May 18th 2008, 06:37 AM

Search Tags


/mathhelpforum @mathhelpforum