Results 1 to 6 of 6

Math Help - Integral

  1. #1
    Member
    Joined
    Jul 2005
    Posts
    187

    Integral

    \int\frac{dx}{e^x+e^{2x}}
    e^x=t
    dx=\frac{dt}{t}
    \int\frac{dt}{t^2(1-t)}
    What's next ?
    Is using partial fractions a good idea ?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,904
    Thanks
    765
    Hello, totalnewbie!

    \int\frac{dx}{e^x+e^{2x}}

    e^x = t\quad\Rightarrow\quad dx = \frac{dt}{t}\quad\Rightarrow\quad \int\frac{dt}{t^2(1-t)} . . . oops! should be 1 + t

    What's next? . Is using partial fractions a good idea?

    Yes . . . an excellent idea!

    Last edited by Soroban; December 15th 2006 at 11:37 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jul 2005
    Posts
    187
    then
    \frac{A}{t^2}+\frac{B}{1-t}=\frac{1}{t^2*(1-t)}<br />
    A=1
    B=1

    then I got:
    \int\frac{dx}{e^x+e^{2x}}=\int t^{-2}+ln(1-t)=-e^{-x}+ln(1-e^x)+ C

    but function calculator gives the following answer: http://wims.unice.fr/wims/wims.cgi?s...on=12&format=t
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by totalnewbie View Post
    then
    \frac{A}{t^2}+\frac{B}{1-t}=\frac{1}{t^2*(1-t)}<br />
    A=1
    B=1

    then I got:
    \int\frac{dx}{e^x+e^{2x}}=\int t^{-2}+ln(1-t)=-e^{-x}+ln(1-e^x)+ C

    but function calculator gives the following answer: http://wims.unice.fr/wims/wims.cgi?s...on=12&format=t
    Why is there a minus, maybe it should be 1+t
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Jul 2005
    Posts
    187
    Quote Originally Posted by ThePerfectHacker View Post
    Why is there a minus, maybe it should be 1+t
    oops, then it's
    \int\frac{dx}{e^x+e^{2x}}=\int t^{-2}+ln(1+t)=-e^{-x}+ln(1+e^x)+ C
    but despite it doesn't match with function calculator answer.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,904
    Thanks
    765
    Hello again, totalnewbie!

    You're missing the "repeated factor" in your Partial Fractions.

    You should have had: . \frac{1}{t^2(1 + t)} \;=\;\frac{A}{t} + \frac{B}{t^2} + \frac{C}{1+t}

    . . and get: . A = -1,\;B = 1,\;C = 1


    Then integrate: . \int\left(-\frac{1}{t} + \frac{1}{t^2} + \frac{1}{1+t}\right)dt \;=\;-\int\frac{dt}{t} + \int t^{-2}dt + \int\frac{dt}{1+t}

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: August 31st 2010, 08:38 AM
  2. Replies: 1
    Last Post: June 2nd 2010, 03:25 AM
  3. Replies: 0
    Last Post: May 9th 2010, 02:52 PM
  4. Replies: 0
    Last Post: September 10th 2008, 08:53 PM
  5. Replies: 6
    Last Post: May 18th 2008, 07:37 AM

Search Tags


/mathhelpforum @mathhelpforum