Partial FractionsDate: 06/08/97 at 15:18:37 From: Grahame Duke Subject: Partial Fractions I am trying to revise for my maths exam next week, but I am having trouble with partial fractions. How does 1/x^2*(x+2) resolve itself into A/2x^2 - 1/4x + 1/4(x+2)? I would have thought that 1/x^2*(x+2) would turn into: A/x + B/x^2 + C/(x+2) I'm trying to solve a problem of integrating between 1 and 2 (x^2+7x+3)/(x^2*(x+3))*dx and I am having trouble resolving the denominator into partial fractions. I think it should be x^2+7x+3 = A/x + B/x^2 + C/(x+3). That means that x^2+7x+3 = A(x)(x+3) + B(x+3) + C(x^2). (This is another process I have never fully grasped!) This means (by my counting) A = 7 (equate coefficients of x), B = 1, and C = -1. Thus I would integrate 7lnx + 1/x - ln(x+3) Unfortunately, my calculator, which can handle integration of this type, doesn't agree with my answer - so I`ve slipped up somewhere. Are there some quick and easy rules regarding partial fractions? My textbook only demonstrates easy problems. Many thanks for any assistance, Grahame Date: 06/09/97 at 17:19:31 From: Doctor Anthony Subject: Re: Partial Fractions Dear Grahame, I will work through the problem, and you can then see where you have made mistakes. x^2+7x+3 A B C Start with: -------- = ----- + ----- + ----- x^2(x+3) x x^2 (x+3) Now if you multiply throughout by the denominator of the lefthand side you will get: x^2+7x+3 = Ax(x+3) + B(x+3) + Cx^2 There are two ways of finding A, B and C. Either multiply out the righthand side and arrange terms in x^2, x and a constant term. Then equate the coefficients of x^2 on each side, equate coefficients of x on each side and finally equate the constant term on both sides. This gives three equations from which you can find A, B and C. A second method is to give x any convenient value, and this will give an equation connecting both sides. Give x another value and get a second equation, and then, if necessary, a third value to get a third equation. Often a mixture of the two methods is the quickest way to find A, B and C. If we use x = 0, then the lefthand side equals 3 and the righthand side equals 3B, so B = 1. Put in x = -3, then the lefthand side equals -9 and the righthand side equals 9C, so C = -1. Finally, the coefficient of x^2 on the left is 1, and on the right A+C, so we can put: A+C = 1 A = 1-C A = 1 - (-1) A = 2 2 1 1 So the fraction can be written ----- + ----- - ----- x x^2 (x+3) -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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