Results 1 to 4 of 4

Math Help - Help please with integration by parts

  1. #1
    Junior Member
    Joined
    Aug 2007
    From
    Houston, TX
    Posts
    46

    Help please with integration by parts

    it says

    Calculate the given integral.
    integral of [x^(7/2) * (ln(3x))] dx

    i started it like this...
    u= ln(3x) du= 1/x dx
    v'= x^(7/2) v= (2/9)x^(9/2)

    =ln(3x)(2/9 x^(9/2)) - 2/9[integral of x^(9/2) * (1/x) dx]

    but now i'm stuck. please help
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    13
    Take it more generally:

    \int x^a\ln(bx)\,dx,\,a>-1,\,b>0.

    The same procedure, integration by parts. Set u=\ln(bx) & dv=x^a\,dx.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Aug 2007
    From
    Houston, TX
    Posts
    46
    sorry, im not sure what you mean
    Follow Math Help Forum on Facebook and Google+

  4. #4
    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 runner07 View Post
    sorry, im not sure what you mean
    you actually did what he said you should do. i don't see why you're stuck

    x^{9/2} \cdot \frac 1x = x^{9/2} \cdot x^{-1} = x^{9/2 - 1} = x^{7/2}

    and you can integrate that using the basic power rule
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: January 11th 2012, 02:30 PM
  2. Replies: 8
    Last Post: September 2nd 2010, 12:27 PM
  3. Replies: 0
    Last Post: April 23rd 2010, 03:01 PM
  4. Integration by Parts!
    Posted in the Calculus Forum
    Replies: 7
    Last Post: January 22nd 2010, 03:19 AM
  5. Replies: 1
    Last Post: February 17th 2009, 06:55 AM

Search Tags


/mathhelpforum @mathhelpforum