Results 1 to 2 of 2

Math Help - Integrate:∫(t-1)ln(t)dt

  1. #1
    Newbie
    Joined
    Nov 2009
    Posts
    21

    Integrate:∫(t-1)ln(t)dt

    ∫(t-1)ln(t)dt


    U=lnt dv/dx = t-1
    Du/dx = 1/t V = ((t^2)/2)-t


    =lnt((t^2/2)-t)- ∫((t^2/2)-t)*1/t


    After doing the calculations I got:
    ((t^2/2)-t)lnt-(t^2/4)+t/2


    The correct answer is
    ((t^2/2)-t)lnt-(t^2/4)+t


    Any help..plz?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member Pranas's Avatar
    Joined
    Oct 2010
    From
    Europe. Lithuania.
    Posts
    81
    Quote Originally Posted by mike789 View Post
    ∫(t-1)ln(t)dt


    U=lnt dv/dx = t-1
    Du/dx = 1/t V = ((t^2)/2)-t


    =lnt((t^2/2)-t)- ∫((t^2/2)-t)*1/t


    After doing the calculations I got:
    ((t^2/2)-t)lnt-(t^2/4)+t/2


    The correct answer is
    ((t^2/2)-t)lnt-(t^2/4)+t


    Any help..plz?
    \displaystyle \[\int {(t - 1)\ln t \cdot {\mkern 1mu} dt} \]

    Notations
    \displaystyle \[u = \ln t \Rightarrow du = \frac{{dt}}{t}\]
    \displaystyle \[dv = (t - 1) \cdot dt \Rightarrow v = \frac{{{t^2}}}{2} - t\]

    Gives us
    \displaystyle \[\int {(t - 1)\ln t \cdot {\mkern 1mu} dt}  = u \cdot v - \int {v \cdot du = } (\frac{{{t^2}}}{2} - t) \cdot \ln t - \int {(\frac{{{t^2}}}{2} - t)} \frac{{dt}}{t} = \]
    \displaystyle \[ = (\frac{{{t^2}}}{2} - t) \cdot \ln t - \int {(\frac{t}{2} - 1)}  \cdot dt = (\frac{{{t^2}}}{2} - t) \cdot \ln t - \frac{{{t^2}}}{4} + t + C\]
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: June 12th 2011, 02:29 PM
  2. Replies: 3
    Last Post: October 8th 2009, 09:58 AM
  3. Replies: 3
    Last Post: July 23rd 2009, 12:33 PM
  4. Replies: 1
    Last Post: December 11th 2008, 01:11 PM
  5. Replies: 3
    Last Post: December 1st 2008, 06:27 PM

Search Tags


/mathhelpforum @mathhelpforum