Results 1 to 3 of 3

Math Help - Indefinite integral help

  1. #1
    Newbie
    Joined
    Oct 2007
    Posts
    16

    Exclamation Indefinite integral help

    S [(x(x-2)/(x-1)^3]

    I can't figure out what u should be.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    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 unluckykc View Post
    S [(x(x-2)/(x-1)^3]

    I can't figure out what u should be.
    first we need to do some manipulation (if we don't do it now, we'd end up doing it later, so let's get it over with).

    Note that \frac {x(x - 2)}{(x - 1)^3} = \frac {x^2 - 2x}{(x - 1)^3} = \frac {x^2 - 2x + 1 - 1}{(x - 1)^3} = \frac {x^2 - 2x + 1}{(x - 1)^3} - \frac 1{(x - 1)^3}

    thus, \int \frac {x(x - 2)}{(x - 1)^3}~dx = \int \frac {x^2 - 2x + 1}{(x - 1)^3}~dx - \int \frac 1{(x - 1)^3}~dx

    now, for the first integral, use the substitution u = (x - 1)^3 for the second integral, use the substitution t = x - 1
    Follow Math Help Forum on Facebook and Google+

  3. #3
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Ah, i just saw it. an easier way is to let u = x - 1 from the beginning. that one substitution will finish the problem

    you would end up with \int \left( \frac 1u - \frac 1{u^3}\right)~du, which should be easy for you

    can you continue?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. indefinite integral
    Posted in the Calculus Forum
    Replies: 1
    Last Post: August 6th 2010, 09:11 PM
  2. Indefinite Integral
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 27th 2010, 06:14 AM
  3. Indefinite integral
    Posted in the Calculus Forum
    Replies: 1
    Last Post: January 8th 2009, 01:50 PM
  4. Indefinite integral
    Posted in the Calculus Forum
    Replies: 2
    Last Post: September 22nd 2008, 03:07 AM
  5. Indefinite integral?
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 15th 2008, 08:36 AM

Search Tags


/mathhelpforum @mathhelpforum