Results 1 to 6 of 6

Math Help - Need help finishing Impartial Fraction

  1. #1
    Member
    Joined
    Apr 2009
    Posts
    85

    Need help finishing Impartial Fraction

    Hello I'm working on an impartial fraction problem

    \int_0^1 \! \frac {x^3-x-3}{x^2-x-6} \, dx.

    Long dividing we get

    \int_0^1 \! x- \frac {x^2-7x-3}{x^2-x-6} \, dx.

    I don't know if i have to long divide again or do partial fraction exp...

    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by mmattson07 View Post
    Hello I'm working on an impartial fraction problem

    \int_0^1 \! \frac {x^3-x-3}{x^2-x-6} \, dx.

    Long dividing we get

    \int_0^1 \! x- \frac {x^2-7x-3}{x^2-x-6} \, dx.

    I don't know if i have to long divide again or do partial fraction exp...

    Thanks

    The division is incorrect:

    [x^3 - x - 3]/[x^2 - x - 6] = x + 1+ [6x + 3] /[x^2 - x - 6] and now do an easy exercise of partial fractions on [6x + 3] /[x^2 - x - 6] = 3(2x + 1)/(x-3)(x+2)

    Tonio
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Apr 2009
    Posts
    85
    Ok so I am stuck again...

    Here is my PFE

    \frac{6x+3}{(x-3)(x+2)}=\frac{A}{x-3}+\frac{B}{x+2}

    6x+3=A(x+2)+B(x-3)

    So A= 21/5 and B= -9/5

    \int_0^1 \! \frac{21}{5}(\frac{1}{x-3})+(\frac{9}{5})\frac{1}{x+2} \, dx.

    \frac{21}{5}ln|x-3|+\frac{9}{5}ln|x+2|\bigg|_0^1

    Im not getting the correct answer...did i make a mistake?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by mmattson07 View Post
    Ok so I am stuck again...

    Here is my PFE

    \frac{6x+3}{(x-3)(x+2)}=\frac{A}{x-3}+\frac{B}{x+2}

    6x+3=A(x+2)+B(x-3)

    So A= 21/5 and B= -9/5 Mr F says: From what I see below I suppose you meant B = 9/5 rather than -9/5.

    \int_0^1 \! \frac{21}{5}(\frac{1}{x-3})+(\frac{9}{5})\frac{1}{x+2} \, dx.

    \frac{21}{5}ln|x-3|+\frac{9}{5}ln|x+2|\bigg|_0^1

    Im not getting the correct answer...did i make a mistake?
    You have forgotten to include \int_0^1 x + 1 \, dx ....
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Apr 2009
    Posts
    85
    So it should be

    \frac{3}{2}+\frac{21}{5}ln(-2)+\frac{9}{5}ln(3)-\frac{21}{5}ln(-3)-\frac{9}{5}ln(2)

    Doesnt look like it will simplify to any of these answers

    \frac{1}{2}-\frac{12}{5}ln(\frac{3}{2})

    \frac{3}{2}-\frac{12}{5}ln(\frac{3}{2})


    \frac{1}{2}-\frac{12}{5}ln(\frac{2}{3})
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by mmattson07 View Post
    So it should be

    \frac{3}{2}+\frac{21}{5}ln(-2)+\frac{9}{5}ln(3)-\frac{21}{5}ln(-3)-\frac{9}{5}ln(2) Mr F says: Things like ln(-2) and ln(-3) are horrible and just plain sloppy. Surely you mean ln|-2| and ln|-3|.

    Doesnt look like it will simplify to any of these answers

    \frac{1}{2}-\frac{12}{5}ln(\frac{3}{2})

    \frac{3}{2}-\frac{12}{5}ln(\frac{3}{2})


    \frac{1}{2}-\frac{12}{5}ln(\frac{2}{3})
    The answer is \frac{3}{2} + \frac{21}{5} \ln \frac{2}{3} + \frac{9}{5} \ln \frac{3}{2}.

    Using a simple log rule this can be re-written as \frac{3}{2} - \frac{21}{5} \ln \frac{3}{2} + \frac{9}{5} \ln \frac{3}{2} ....
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. help finishing of a question
    Posted in the Algebra Forum
    Replies: 4
    Last Post: February 19th 2010, 07:01 AM
  2. Help finishing a proof
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: October 21st 2009, 09:33 PM
  3. I need help finishing this problem.
    Posted in the Calculus Forum
    Replies: 8
    Last Post: July 29th 2009, 01:54 PM
  4. [SOLVED] HELP! I need help finishing this problem!
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: February 12th 2008, 08:12 PM
  5. Can I get help finishing this logerithm?
    Posted in the Algebra Forum
    Replies: 2
    Last Post: November 28th 2007, 01:22 AM

Search Tags


/mathhelpforum @mathhelpforum