Results 1 to 4 of 4

Math Help - patrial fraction decomp integration

  1. #1
    Junior Member
    Joined
    Sep 2008
    Posts
    30

    patrial fraction decomp integration

    can anyone help me do this integration
    (dx)/(x^3 - 2x^2 + 4x - 8)

    I factored the bottom to (x^2 + 4)(x - 2)

    and found
    A = -1/8
    B = 1/4
    C = 1/8

    don't really know what to do next
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Oct 2005
    From
    Earth
    Posts
    1,599
    Assuming those numbers are correct, the next thing to do is to separate the original fraction into two separate ones, both which can be integrated separately.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Sep 2008
    Posts
    30
    hmm

    the integral of

    (1/4) / (x^2 + 4)

    is this (1/4)arctan(x/2) ?
    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 sleepiiee View Post
    hmm

    the integral of

    (1/4) / (x^2 + 4)

    is this (1/4)arctan(x/2) ?
    No.

    \frac{\frac{1}{4}}{x^2 + 4} = \frac{1}{8} \cdot \frac{2}{x^2 + 2^2} \, ....
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Patrial fractions?
    Posted in the Calculus Forum
    Replies: 2
    Last Post: August 31st 2010, 12:42 AM
  2. partial-fraction-decomp question
    Posted in the Pre-Calculus Forum
    Replies: 8
    Last Post: July 20th 2010, 06:39 PM
  3. Partial Fraction Decomp Integral
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 4th 2010, 01:20 PM
  4. Partial Fraction Decomp Help!
    Posted in the Calculus Forum
    Replies: 2
    Last Post: December 3rd 2008, 04:12 PM
  5. Help with Partial Fraction Decomp
    Posted in the Algebra Forum
    Replies: 1
    Last Post: December 9th 2007, 07:05 AM

Search Tags


/mathhelpforum @mathhelpforum