Results 1 to 6 of 6

Math Help - Integral of ratio

  1. #1
    Ant
    Ant is offline
    Member
    Joined
    Apr 2008
    Posts
    137
    Thanks
    4

    Integral of ratio

    Hi,



    Can anyone explain how this follows? Has it been factories and simplified?

    Thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Santa Cruz, CA
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by Ant View Post
    Hi,



    Can anyone explain how this follows? Has it been factories and simplified?

    Thanks!
    Since you're integrating an improper fraction, we need to first make it proper.

    By long division, you should see that \dfrac{2t^3}{1+t}=2t^2-2t+2-\dfrac{2}{t+1}

    Now integrate this result instead. Can you proceed?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5
    Is...

    \displaystyle \frac{2\ t^{3}}{1+t} = 2\ t^{2} - \frac{2\ t^{2}}{1+t} = 2\ t^{2} - 2\ t + \frac{2\ t}{1+t} = 2\ t^{2} - 2\ t + 2 - \frac{2}{1+t}

    Kind regards

    \chi \sigma
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Ant
    Ant is offline
    Member
    Joined
    Apr 2008
    Posts
    137
    Thanks
    4
    Yes thanks, it's all quite straight forward from there. I think I need to revisit algebraic long division though it's been about 3 years since I've used it!

    RE the method used directly above, in general when given:

    A/B = (C.B)/B - D/B = C - D/B ???

    This bears some resemblance to completely the square?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Mar 2010
    Posts
    715
    Thanks
    2
    Here is a trick: let y = 1+t so that y-1 = t, then:

    \begin{aligned} & \frac{2t^3}{1+t} = \frac{2(y-1)^3}{y} = \frac{2 y^3-6 y^2+6 y-2}{y} = \\& 2y^2-6y+6-\frac{2}{y} = 2(1+t)^2-6(1+t)\\&+6-\frac{2}{1+t} = 2t^2-2 t+2-\frac{2}{t+1}.\end{aligned}

    Or, better yet, use the substitution as an integral substitution.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Santa Cruz, CA
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by TheCoffeeMachine View Post
    Here is a trick: let y = 1+t so that y-1 = t, then:

    \begin{aligned} & \frac{2t^3}{1+t} = \frac{2(y-1)^3}{y} = \frac{2 y^3-6 y^2+6 y-2}{y} = \\& 2y^2-6y+6-\frac{2}{y} = 2(1+t)^2-6(1+t)\\&+6-\frac{2}{1+t} = 2t^2-2 t+2-\frac{2}{t+1}.\end{aligned}

    Or, better yet, use the substitution as an integral substitution.
    Or even better,

    \begin{aligned}\dfrac{2t^3}{t+1} &= \dfrac{2(t^3+1)}{t+1}-\dfrac{2}{t+1}\\ &= \dfrac{2(t+1)(t^2-t+1)}{t+1}-\dfrac{2}{t+1}\\ &= 2t^2-2t+2-\dfrac{2}{t+1}\end{aligned}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. integral of ratio of cosine and polynomial
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 21st 2011, 12:48 PM
  2. Ratio for curve and point-ratio form
    Posted in the Algebra Forum
    Replies: 4
    Last Post: January 11th 2010, 12:07 AM
  3. Finding n given the ratio of nCr to another ratio
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: October 5th 2009, 07:03 PM
  4. Ratio help
    Posted in the Algebra Forum
    Replies: 2
    Last Post: April 29th 2009, 08:04 AM
  5. Ratio
    Posted in the Algebra Forum
    Replies: 6
    Last Post: April 10th 2007, 03:55 PM

Search Tags


/mathhelpforum @mathhelpforum