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Homework Help: Partial fraction Urgent

  1. Jun 14, 2005 #1
    Hi every one.

    Just came acroos a nasty piece and was wondering if you could help me with it.

    I wasn't sure if it belong to math section or here but scince it is for my homework I placed it here.
    Here is the question.
    Find the integral of [tex]\frac{x}{(x+1)(x-2)^2}[/tex]
    I have gotten to some piont using the partial fraction.
    here is my work so far.

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

    I found the values of A B and C.

    C=[tex]\frac{2}{3}[/tex] B=[tex]\frac{2}{3}[/tex] A=[tex]\frac{21}{9}[/tex]

    If you could be so kind and help me through with it by typing what you have done would be really nice, I am not asking in latex form a simple typing will do the job.

    Also can anyone help me finding the equation of the graph attached?
     

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    Last edited: Jun 14, 2005
  2. jcsd
  3. Jun 14, 2005 #2
    No-one is going to do your work for you. You've done the partial fractions, so what's wrong with doing the integration?
     
  4. Jun 14, 2005 #3
    If you've found A B and C ccorrectly (havent checked yet) then you can just substitue your new expression into hte integral.

    [tex] \int \frac{x}{(x+1)(x-2)^2} dx = \int \left(\frac{2}{3(x+1)} + \frac{2}{3(x-2)} + \frac{7}{3(x-2)}\right) dx [/tex] which are all trivial integrals.

    If those are correct values for A B and C it simplifies even more.
     
  5. Jun 14, 2005 #4
    your partial fractions is wrong. i get this system

    A+B=0
    -4A-B+C=1
    4A-2B+C=0

    then solve for a b and c again. then plug them in and replace your initial integral with your partial fractions. shouldn't be too bad from there. show your work if you're still stuck.
     
  6. Jun 14, 2005 #5
    Fine Gale, take the kill.
     
  7. Jun 14, 2005 #6
    Let me first check if I have it right.

    To get partial fraction I would need 3 definitions, right?

    Like

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

    Then solve for A,B and C.

    then do the integration.

    In my last attempt I ended up with a [tex]Ln[/tex] and a function.

    Is it sposed to be like that?

    Any help with the graph?
     
  8. Jun 14, 2005 #7
    you should end up with two Ln functions and a rational function after the integration. plus don't forget an integrating constant.

    The graph hasn't been approved yet, so can't help you till we see it.
     
  9. Jun 14, 2005 #8
    Thats the correct equation, equate it to the original integrand and multiply out by the denominator, and solve the resulting system of equations for A B and C. Once you get those, itnegrate the equivalent partial fractions, each one will evaluate to a LN of a function except the last one.
     
  10. Jun 14, 2005 #9
    Thanx guys.
     
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