1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Set up Partial Fractions

  1. May 31, 2009 #1
    1. The problem statement, all variables and given/known data

    I need to intergrate the following

    (4x2+2x-1)/ (x3+x2)
    How to set up problem using linear factors?

    3. The attempt at a solution

    Factoring the denominator I get:
    x2(x+1)

    By linear/quadratic factoring I get:
    A/x + Bx+C/ x2 + d/(x+1)

    Is this right?
     
  2. jcsd
  3. May 31, 2009 #2
    Correct. All you need to do is to solve for the coefficients.

    If you don't like to do that, you can use the following alternative method. We have:

    R(x) =(4x^2+2x-1)/ (x^3+x^2) =

    (4x^2+2x-1)/ [x^2(1+x)]

    Near x = -1 this is:

    R(x) = 1/(1+x) * {[4x^2 + 2 x - 1]/x^2 at x = -1} =

    1/(1+x)

    Now, a more accurate expansion of R(x) near x = 1 would be obtained by multiplying 1/(1+x) by the expansion of [4x^2 + 2 x - 1]/x^2 around the point x = 1, then you would get an expansion of the form:

    1/(1+x) [1 + A (1+x) + B (1+x)^2 + ...] =

    1/(1+x) + A + B(1+x)^2


    But only the first term in ths expansion is singual, it diverges as you let x tend to 1.

    If we now consider the expansion of R(x) near the other singular point, x = 0, and collect all the singular terms, we get:

    (4x^2+2x-1)/ [x^2(1+x)] =

    1/x^2 [4x^2 + 2x - 1] series expansion of 1/(1+x) =


    1/x^2 [4 x^2 + 2 x - 1] [1-x + x^2 - ....]=

    -1/x^2 + 3/x + nonsingular terms.

    So, we have that near x = 1, the singular behaviour of R(x) is given by 1/(1+x) while near x = 0 it is given by -1/x^2 + 3/x. Then it follows that the function:

    R(x) - [1/(1+x) - 1/x^2 + 3/x]

    is a rational function that has no singularities because the only singular points of R(x) are at zero and at x = 1 but we have subtracted the singular terms of R around these points. It thus follws that:

    R(x) - [1/(1+x) - 1/x^2 + 3/x]

    is a polynomial. But since both R(x) and the subtracted terms tend to zero at infinity this polynomial must be identical to zero. Therefore we have:

    R(x) - [1/(1+x) - 1/x^2 + 3/x] = 0 ---------->

    R(x) = 1/(1+x) - 1/x^2 +3/x

    And we didn't need to solve a single equation to get this result.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Set up Partial Fractions
  1. Partial fractions (Replies: 3)

  2. Partial fractions (Replies: 4)

  3. Partial fractions (Replies: 2)

Loading...