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and the problem reduces to nding the antiderivative of a proper rational function.
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The theorems that follow hold for polynomials with arbitrary real coef cients. However, for simplicity we shall illustrate them only with polynomials whose coef cients are integers.
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Copyright 2008, 1997, 1985 by The McGraw-Hill Companies, Inc. Click here for terms of use.
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THE METHOD OF PARTIAL FRACTIONS
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[CHAP. 40
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x2 + 1 x 2 1 x 4 + 7x x4 x2 x 2 + 7x x2 1 7x + 1
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Fig. 40-1 Theorem 40.1: Any polynomial D(x) with leading coef cient 1 can be expressed as the product of linear factors, of the form x a, and irreducible quadratic factors (that cannot be factored further), of the form x 2 + bx + c, repetition of factors being allowed. As explained in Section 7.4, the real roots of D(x) determine its linear factors. EXAMPLES
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(a) (b) x 2 1 = (x 1)(x + 1) Here, the polynomial has two real roots ( 1) and, therefore, is a product of two linear factors. x 3 + 2x 2 8x 21 = (x 3)(x 2 + 5x + 7) The root x = 3, which generates the linear factor x 3, was found by testing the divisors of 21. Division of D(x) by x 3 yielded the polynomial x 2 + 5x + 7. This polynomial is irreducible, since, by the quadratic formula, its roots are x= which are not real numbers. b 5 3 b2 4c = 2 2
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Theorem 40.2 (Partial Fractions Representation): Any (proper) rational function f (x) = N(x)/D(x) may be written as a sum of simpler, proper rational functions. Each summand has as denominator one of the linear or quadratic factors of D(x), raised to some power. By Theorem 40.2, f (x)dx is given as a sum of simpler antiderivatives antiderivatives which, in fact, can be found by the techniques already known to us. It will now be shown how to construct the partial fractions representation and to integrate it term by term. Case 1: D(x) is a product of nonrepeated linear factors. The partial fractions representation of f (x) is N(x) A1 A2 An + + + = (x a1 )(x a2 ) (x an ) x a1 x a2 x an The constant numerators A1 , . . . , An are evaluated as in the following example. EXAMPLE
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A1 A2 2x + 1 = + (x + 1)(x 1) x+1 x 1 A A 2x + 1 = (x + 1)(x 1) 1 + (x + 1)(x 1) 2 (x + 1)(x 1) x+1 x 1 2x + 1 = A1 (x 1) + A2 (x + 1) In (1), substitute individually the roots of D(x). With x = 1, 1 = A1 ( 2) + 0 or A1 = 1 2 (1)
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Clear the denominators by multiplying both sides by (x + 1)(x 1), (x + 1)(x 1)
CHAP. 40]
THE METHOD OF PARTIAL FRACTIONS
and with x = 1, 3 2 With all constants known, the antiderivative of f (x) will be the sum of terms of the form 3 = 0 + A2 (2) or A2 = A dx = A ln |x a| x a
Case 2: D(x) is a product of linear factors, at least one of which is repeated. This is treated in the same manner as in Case 1, except that a repeated factor (x a)k gives rise to a sum of the form A2 Ak A1 + + + 2 x a (x a) (x a)k EXAMPLE
Multiply by (x l)2 (x 2), 3x + 1 = A1 (x 1)(x 2) + A2 (x 2) + A3 (x 1)2 Letting x = 1, 4 = 0 + A2 ( 1) + 0 Letting x = 2, 7 = 0 + 0 + A3 (1) or A3 = 7 The remaining numerator, A1 , is determined by the condition that the coef cient of x 2 on the right side of (2) be zero (since it is zero on the left side). Thus, A 1 + A3 = 0 or A1 = A3 = 7 or A2 = 4 (2) A3 3x + 1 A1 A2 + + = x 1 (x 1)2 x 2 (x 1)2 (x 2)
[More generally, we use all the roots of D(x) to determine some of the A s, and then compare coef cients of as many powers of x as necessary to nd the remaining A s.]
Now the antiderivatives of f (x) will consist of terms of the form ln |x a| plus at least one term of the form A B dx = j (x a) (x a) j 1 Case 3: D(x) has irreducible quadratic factors, but none is repeated. In this case, each quadratic factor x 2 + bx + c contributes a term x2 to the partial fractions representation. EXAMPLE
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