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I need help integrating fractions

  1. Mar 5, 2005 #1
    We just started this and I mostly understand it except when it comes to using A, B, C, etc substitution. What I mean is this, here is an exampe.

    (6x^2+x+1)/(x^2+1)(x-1) = (Ax+B)/(x^2+1) + (C)/(x-1)

    You then multiply by denominators so you end up with (Ax+B)*(x-1) + (C)*(x^2+1). You multiply and factor and end up with (A+C)x + (-A+B)x + (-B+C). Then you solve the equations for A B and C since (A+C)=6, (-A+B)=1 and (-B+C)=1.

    But how do you know which denominators to multiply the numerators by? In some problems you end up with 5 or 6 different denominators, and I do not know which ones to multiply the denominators by. In some problems (unlike the one above) it is broken down into 5 or 6 different integrals instead of 2 and if I don't know which ones to multiply out I cannot solve for A, B, C, etc.
  2. jcsd
  3. Mar 5, 2005 #2
    You multiply through by the entire denominator. Remember that it's an equation, so the two sides must be equal. In your example,

    [tex]\frac{(6x^2+x+1)}{(x^2+1)(x-1)} = \frac{(Ax+B)}{(x^2+1)} + \frac{C}{(x-1)} [/tex]

    you'd multiply both sides by (x2 + 1)(x-1), to give you

    6x2 + x + 1 = (Ax + B)(x - 1) + C(x2 + 1),

    and you would then expand out the right side and match coefficients of powers of x, as you did before (that's where those equtions come from!) or you could substitute in some choice values of x to easily solve for some variables.

    For instance, if you let x = 1, you get the equation

    8 = (A + B)*(0) + 2C
    8 = 2C
    C = 4

    You can't pull this trick for all of the values, but it's good to get a few easy ones.

  4. Mar 5, 2005 #3


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    With all:
    [tex] \frac{A}{Bx+C}+\frac{D}{Ex+F}+\frac{G}{Hx+I}+\frac{J}{Kx+L}+...[/tex]

    In this simple example,you have to multiply A with all the other monoms,except "Bx+C"...

  5. Mar 6, 2005 #4
    My calculus professor had this weird thing he does that even when I asked him about it he didn't explain it in a way that I understood it. It was

    Ax+B/(x^2+1) + Cx+D/(x^2+1)^2 + Ex+F/(x^2+2x-2) + G/(x+1) + H/(x+1)^2 + I/(x-2)

    he wouldn't just multiply Ax+B by all of them, he may exclude (as an example) the denominator from G and I have no idea what he was doing or why when he did that.

    For the record, can I just multiply out by all the denominators and get the right answers?
  6. Mar 6, 2005 #5


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    Ooops,you're right,my bad.Yes,multiplication only with the prime polynomials.Of course,the repeated ones,are not multiplicated twice...

  7. Mar 6, 2005 #6
    What are the 'repeated ones' and what constitutes a 'repeated one'
  8. Mar 6, 2005 #7


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    [tex] \frac{x+7}{(x-3)^2}=\frac{A}{x-3}+\frac{Bx+C}{(x-3)^{2}}[/tex]

    You see,A is multiplied not with the other denominator...

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