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Math Help - Partial fraction

  1. #1
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    Partial fraction

    \int \frac {x^2}{(x-3)(x+2)^2} ~dx = \frac {A}{x-3} + \frac {B}{x+2} + \frac {C}{(x+2)^2}

    That is where I am at so far, and I just need a little help with finding the common denominator. Normally you would multiply A by the denominators of both B and C, but what about in this case? Since they are practically the same. Would I multiply A by (x+2)(x+2)^2 or just by (x+2)?
    Last edited by redman223; July 6th 2008 at 10:43 PM.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by redman223 View Post
    \int \frac {x^2}{(x-3)(x+2)^2} ~dx = \frac {A}{x-3} + \frac {B}{x-2} + \frac {C}{(x-2)^2}

    That is where I am at so far, and I just need a little help with finding the common denominator. Normally you would multiply A by the denominators of both B and C, but what about in this case? Since they are practically the same. Would I multiply A by (x-2)(x-2)^2 or just by (x-2)?
    the common denominator is always the denominator of the original fraction. are you sure it should be x - 2 or x + 2?
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    Oops, it was supposed to be x+2, it has been fixed. So how are those treated?
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    Hello,

    Multiply the whole by (x+2)^2, but by (x-3)(x+2)^2 would be more straightforward :

    x^2=A(x+2)^2+B(x+2)(x-3)+C(x-3)

    Then, let x=-2 to annulate x+2 ---> you will have C
    After that, let x=3 to annulate x-3 ---> you will have A
    And then, pick up a random value for x, for example x=0, to get B.
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    How does setting x = 0 allow you to find B?
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  6. #6
    Senior Member nikhil's Avatar
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    Lightbulb See the basics

    Quote Originally Posted by redman223 View Post
    How does setting x = 0 allow you to find B?
    Actually didn't understood what you actually meant.
    Suppose
    (x+1)/(x+2)(x+3)=[A/(x+2)] +B/x+3 or we may say
    (x+1)/(x+2)(x+3)=[A(x+3)+B(x+2)]/(x+2)(x+3) or
    x+1=A(x+3)+B(x+2)
    if you put x=-2 you will get
    -1=A(-2+3)+B(-2+2) or
    -1=A
    or A=-1
    we take such value of x for which atleast 1 expression may become 0 so that calculating other constant may be calculated easily.
    NOTE:actually same thing can be done by comparing LHS and RHS. But this substitution method is much easier.
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  7. #7
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    Quote Originally Posted by redman223 View Post
    How does setting x = 0 allow you to find B?
    Once you find the value for A and C, you will only have B as an unknown in the equation (left to found) when x=0 or any value you choose because you can insert values for A and C.
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