Results 1 to 4 of 4

Math Help - definite integral (logarithmic) completely lost

  1. #1
    Junior Member
    Joined
    Sep 2010
    Posts
    60

    definite integral (logarithmic) completely lost

    problem:

    Integrate (2x-1)/(x+1) from 0 to 3

    the first thing I do is pull out the constant 2. Then the part to integrate can become ln(x-1)-ln(x+1)

    the antiderivative of ln(x-1) is:
    x(ln(x-1)-1)-ln(1-x)

    the antiderivative of ln(x+1) is:
    (x+1)ln(x+1)-x

    x(ln(x-1)-1)-ln(1-x)-(x+1)ln(x+1)-x

    3(ln(3-1)-1)-ln(1-3)-(3+1)ln(3+1)-x

    I can already tell this answer is wrong due to the fact that I have negatives in the ln

    the final answer should be 6(1-ln2)

    any help?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Mar 2012
    From
    Unknown
    Posts
    29
    Thanks
    3

    Re: definite integral (logarithmic) completely lost

    Quote Originally Posted by NecroWinter View Post
    Then the part to integrate can become ln(x-1)-ln(x+1)
    How? I've no idea how you have arrived at this step.

    To do your integral note (2x-1)/(x+1) = 2-[3/(x+1)].
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Sep 2010
    Posts
    60

    Re: definite integral (logarithmic) completely lost

    Quote Originally Posted by TheSaviour View Post
    How? I've no idea how you have arrived at this step.

    To do your integral note (2x-1)/(x+1) = 2-[3/(x+1)].
    its possible I confused the property of logarithms.

    Anyway, how do you go from (2x-1)/(x+1) to 2-[3/(x+1)]
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Mar 2012
    From
    Unknown
    Posts
    29
    Thanks
    3

    Re: definite integral (logarithmic) completely lost

    [(2x-1)/(x+1)] =[(2x+2-3)/(x+1)] = [2(x+1)-3]/(x+1) = 2(x+1)/(x+1)-[3/(x+1)] = 2-[3/(x+1)].
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Completely Lost in Translation/Simplification
    Posted in the Algebra Forum
    Replies: 4
    Last Post: January 4th 2010, 04:15 PM
  2. im completely lost
    Posted in the Algebra Forum
    Replies: 1
    Last Post: November 15th 2009, 10:39 AM
  3. Replies: 11
    Last Post: June 30th 2008, 08:26 PM
  4. i am completely lost!
    Posted in the Algebra Forum
    Replies: 1
    Last Post: October 11th 2007, 04:56 AM
  5. I'm Completely Lost
    Posted in the Algebra Forum
    Replies: 4
    Last Post: September 12th 2007, 07:29 PM

Search Tags


/mathhelpforum @mathhelpforum