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!

Help with Partial Fraction Decomposition

  1. Jun 26, 2010 #1
    1. The problem statement, all variables and given/known data

    [tex]\int[/tex] 1/(x[tex]^{2}[/tex]+4x+5)[tex]^{2}[/tex]

    2. Relevant equations

    I am using partial fraction decomposition

    3. The attempt at a solution

    1/(x[tex]^{2}[/tex]+4x+5)[tex]^{2}[/tex] = Ax+b/(X^2+4x+5) + Cx+D/(x^2+4x+5)^2

    1 = (Ax+b)(X^2+4x+5) + Cx+D

    When i multiply through to find the values for A and b and c i get zero because there is no X^3 or x^2 or x on the left side. Am i doing something wrong or do i need to use another method to solve this integral?
     
  2. jcsd
  3. Jun 26, 2010 #2

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    The integrand is already in expanded form. You need to use another method. Start by rewriting the denominator in the form (x+a)2+b2, and then use a trig substitution.
     
  4. Jun 26, 2010 #3
    so when there is only dx sitting on the top i cannot use partial fraction decomposition? How can i tell if its already in expanded form. On the directions of this problem the TA said to use PFD but it clearly does not work.
     
  5. Jun 26, 2010 #4

    Mark44

    Staff: Mentor

    If you use partial fraction decomposition, I'm reasonably certain that you'll end up with exactly what you started with. Take vela's advice.
     
  6. Jun 26, 2010 #5

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    You want to use partial fractions when you have distinct factors on the bottom. For example, you would use the method if the denominator was (x+1)(x2+1). On the other hand, if you had just (x+1)2 in the denominator, it's already in the form you want it in. For instance, if you tried to expand, 1/(x+1)2, you'd write

    [tex]\frac{1}{(x+1)^2} = \frac{A}{x+1} + \frac{B}{(x+1)^2}[/tex]

    But the LHS already looks just like the second term on the RHS with B=1. Similarly, in this problem, the integrand is already in the form of a possible term you'd have after doing the partial fractions expansion, which is why you're just ending up with what you started with.

    Your TA probably just glanced at the problem and told you to use partial fractions before thinking about it. When you see that type of integrand, partial fractions is one of the first methods that comes to mind.
     
  7. Jun 26, 2010 #6
    oh, i see what you mean. Basically if there is only one type of polynomial squared or cubed in the denominator then i cant use partial fraction but if suppose there is (x+2)(x-1)^2 in the denominator then i could use partial fractions and expand the as a/(x+2) + b/(x-1) + c/(x-1)^2 right? Also, in order to solve the first question i posted should i complete the square for the denominator in order to get it into a trig substitution format?
     
  8. Jun 26, 2010 #7

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Yes and yes.
     
  9. Jun 27, 2010 #8
    After i complete the square i end up with dx/((x+2)^2+1)^2. I substitute u = x+2 , du = dx
    so i get the integral du/(u^2+1)^2. Then i got u = tan theta, du = sec^2(theta) d(theta) and i replace that in the integral as sec^2(theta) d(theta)/(tan^2(theta) +1)^ and i get after i substitute sec^2 in the bottom for tan^2 + 1 and then cancel the sec^2 on the top i end up with 1/sec^2, which is cos^2. Am i on the right path so far?
     
  10. Jun 27, 2010 #9

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Yup!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Help with Partial Fraction Decomposition
Loading...