Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: What is wrong with this method? (partial fractions)

  1. Jan 31, 2012 #1
    1. The problem statement, all variables and given/known data

    So I am trying to integrate the following. I believe I have to use partial fractions.


    2. Relevant equations

    3. The attempt at a solution

    So I am trying to expand the above into something that I can integrate. So I equated the integrand with the following:

    integrand = [itex]\frac{A}{(x-3)}[/itex]+[itex]\frac{Bx+C}{(x^{2}+16+8x) <--just the expanded form of (x+4)^{2}}[/itex]

    So by creating a common denominator you get

    1= A(x[itex]^{2}[/itex]+16+8x)+(Bx+C)(x-3)

    By inputing x=3 first, I solve for A=1/49
    Then by imputing x=0 I get C=-33/49
    And then by imputing x=1 I get B=-45/49

    So plugging that into the original, I get:

    integrand = [itex]\frac{1}{49(x-3)}[/itex]-[itex]\frac{(45/49)x-(33/49)}{(x+4)(x+4)}[/itex]

    When I check my answer by plugging a random value of x into the integrand, and then into the RHS, I find that they are not equal. Where did I mess up?
  2. jcsd
  3. Jan 31, 2012 #2


    User Avatar
    Science Advisor

    I don't know why you multiplied out [itex](x+ 4)^2[/itex]. If you leave it as is, you have
    [tex]\frac{1}{(x+4)^2(x-3)}= \frac{A}{x-3}+ \frac{Bx+ C}{(x+4)^2}[/tex]
    Multiplying on both sides by [itex](x-3)(x+4)^2[/itex] we get
    [tex]1= A(x+4)^2+ (Bx+ C)(x- 3)[/tex]

    Now, you can take x= 3 and get [itex]1= A(7)^2[/itex] so A= 1/49 as you say.

    If you let x= 0 you get [itex]1= 16A+ C(-7)[/itex] and knowing that A= 1/49, you get 7C= 16/49- 1= -33/49 so C= -33/343, not -33/49.

    Finally, if you let x=-4, you get [itex]1= (-4B+ C)(-7)[/itex] and can solve for B.
  4. Jan 31, 2012 #3
    OK Thanks for the help!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook