Results 1 to 5 of 5

Math Help - Integration By Parts

  1. #1
    Member
    Joined
    Nov 2008
    Posts
    76

    Integration By Parts

    Use integration by parts to find the integral:
    Attached Thumbnails Attached Thumbnails Integration By Parts-question-4.jpg  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor kalagota's Avatar
    Joined
    Oct 2007
    From
    Taguig City, Philippines
    Posts
    1,026
    Quote Originally Posted by My Little Pony View Post
    Use integration by parts to find the integral:
    for the first time, let u = (\ln (4x))^2 and dv=dx

    and for the second time, let u= \ln 4x and dv=dx
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie Mr. Engineer's Avatar
    Joined
    Apr 2008
    From
    Chicago, Illinois
    Posts
    24

    Reply

    Integration by parts;

    \int u \, dv=uv - \int v \, du

    Function:
    \int(ln(4x))^2

    let u=(ln(4x))^2
    dv=dx

    du=2(ln(4x))\frac{1}{4x}
    v=x

    = x(ln(4x))^2-\int2x(ln(4x))\frac{1}{4x}

    Then simplify the remaining integral and integrate once again
    Last edited by mr fantastic; February 19th 2009 at 03:45 AM. Reason: Fixed the first line of latex
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Nov 2008
    Posts
    76
    Quote Originally Posted by Mr. Engineer View Post
    Integration by parts;

    \int u \, dv=uv - \int v \, du

    Function:
    \int(ln(4x))^2

    let u=(ln(4x))^2
    dv=dx

    du=2(ln(4x))\frac{1}{4x}
    v=x

    = x(ln(4x))^2-\int2x(ln(4x))\frac{1}{4x}

    Then simplify the remaining integral and integrate once again
    Are you sure it's 1/4x and not 1/x?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie Mr. Engineer's Avatar
    Joined
    Apr 2008
    From
    Chicago, Illinois
    Posts
    24

    reply

    my bad it is 1/x
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: January 11th 2012, 03:30 PM
  2. Replies: 8
    Last Post: September 2nd 2010, 01:27 PM
  3. Replies: 0
    Last Post: April 23rd 2010, 04:01 PM
  4. Integration by Parts!
    Posted in the Calculus Forum
    Replies: 7
    Last Post: January 22nd 2010, 04:19 AM
  5. Replies: 1
    Last Post: February 17th 2009, 07:55 AM

Search Tags


/mathhelpforum @mathhelpforum