Results 1 to 8 of 8

Math Help - integration question

  1. #1
    Newbie
    Joined
    Apr 2008
    Posts
    1

    integration question

    how do you integrate y= e^(-3x)Ln(x)

    ive tried integration by parts, but i cant find a way that works.

    I'm sure theres a very simple way and i just cant see it.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Apr 2008
    Posts
    72
    to integrate e divide by the derivative of the power, leave the power the same..

    the power is (-3x)Ln(x) which is differentiated using the product rule f'g+g'f
    f=-3x f'=-3 g=ln(x) g'=1/x the differential of the power is -3ln(x)+-3x/x which simplifies to -3ln(x)-3

    The integral of y= e^(-3x)Ln(x) is therfore y= (e^(-3x)Ln(x))/(-3ln(x)-3)+c
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Bar0n janvdl's Avatar
    Joined
    Apr 2007
    From
    South Africa
    Posts
    1,630
    Thanks
    6
    Quote Originally Posted by robertsays View Post
    how do you integrate y= e^(-3x)Ln(x)

    ive tried integration by parts, but i cant find a way that works.

    I'm sure theres a very simple way and i just cant see it.
    Just to clarify things, do you mean:

     \int e^{-3x \ln x}

    or

     \int e^{-3x} \ln x


    EDIT: Mathematica Integrator says the first one most probably doesn't exist... (When first editing my post I accidentally said the second one doesn't exist. My apologies.)
    Last edited by janvdl; April 6th 2008 at 01:21 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Apr 2008
    Posts
    72

    is the 2nd option possible?

    I haven't learnt any way to integrate products other than ones such as trigonometric products where there is a way to get a sum...

    Also he hasn't put a x or * to indicate that its e(-3x) times ln(x) so i htink its safe to assume he/she means e(-3x*ln(x))
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Bar0n janvdl's Avatar
    Joined
    Apr 2007
    From
    South Africa
    Posts
    1,630
    Thanks
    6
    Quote Originally Posted by deragon999 View Post
    I haven't learnt any way to integrate products other than ones such as trigonometric products where there is a way to get a sum...

    Also he hasn't put a x or * to indicate that its e(-3x) times ln(x) so i htink its safe to assume he/she means e(-3x*ln(x))
    Mathematica's Integrator finds an answer for the 2nd one... But I cannot find it myself. I tried partial integration.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by deragon999 View Post
    to integrate e divide by the derivative of the power, leave the power the same..

    the power is (-3x)Ln(x) which is differentiated using the product rule f'g+g'f
    f=-3x f'=-3 g=ln(x) g'=1/x the differential of the power is -3ln(x)+-3x/x which simplifies to -3ln(x)-3

    The integral of y= e^(-3x)Ln(x) is therfore y= (e^(-3x)Ln(x))/(-3ln(x)-3)+c
    your method does not work in general and is therefore not correct. in particular, it is not correct in this case (just check by differentiating your answer).

    either interpretation yields a nasty integral in this case. one interpretation yields a solution in non-elementary terms, the other can't even be found by the integrator, so it's probably not integrable
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member angel.white's Avatar
    Joined
    Oct 2007
    Posts
    723
    Awards
    1
    Both involve what I assume is Big-O notation.
    Attached Thumbnails Attached Thumbnails integration question-crazyintegral1.jpg   integration question-crazyintegral2.jpg  
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Bar0n janvdl's Avatar
    Joined
    Apr 2007
    From
    South Africa
    Posts
    1,630
    Thanks
    6
    Quote Originally Posted by angel.white View Post
    :/
    angel.white, try this integrator too: The Integrator--Integrals from Mathematica
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. integration question
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 30th 2011, 07:19 AM
  2. Replies: 6
    Last Post: July 21st 2010, 06:20 PM
  3. Integration Question
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 20th 2010, 02:59 AM
  4. help with integration question?
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 19th 2010, 02:49 PM
  5. integration question no. 7
    Posted in the Calculus Forum
    Replies: 2
    Last Post: November 11th 2007, 08:40 AM

Search Tags


/mathhelpforum @mathhelpforum