Steps for Integration by Partial Fractions

Steps for Integration by Partial Fractions - ( 29 k x c-,...

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Steps for Integration by Partial Fractions Problem Statement : Evaluate ( 29 ( 29 p x dx q x , where p and q are polynomials with no common factors. So ( 29 1 1 0 n n n n p x a x a x a - - = + + + L , 0 n a , and ( 29 1 1 0 m m m m q x b x b x b - - = + + + L , 0 m b . Step 0 : If n m , divide the fraction out (using long division), obtaining a quotient Q and a remainder R for which ( 29 ( 29 ( 29 ( 29 ( 29 p x R x Q x q x q x = + , and the degree of R is smaller than the degree of q . Notice that Q is a polynomial and therefore easy to integrate. The problem is how to integrate ( 29 ( 29 R x q x . Partial Fractions is a technique that applies to this situation. Step 1 : Factor the denominator completely. For factors that are irreducible quadratics, complete the square so that they are of the form ( 29 2 2 k x c d - + . Of course, linear factors are of the form ( 29 k x c - . Step 2 : Decompose the integrand into the most general sum of fractions. For each linear factor in the denominator of the form
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Unformatted text preview: ( 29 k x c-, there will be a sum of the form ( 29 ( 29 1 1 1 k k k k A A A x c x c x c--+ + +---L . For each irreducible quadratic factor in the denominator of the form ( 29 2 2 k x c d -+ , there is a sum of the form ( 29 ( 29 ( 29 1 1 1 1 1 2 2 2 2 2 2 k k k k k k B x C B x C B x C x c d x c d x c d---+ + + + + +-+ -+-+ L . The A s, B s, and C s are unknown constants. Step 3 : Clear the fractions in Step 2 and solve for the unknown constants. Step 4 : Using the results of Step 3, replace the integrand with the partial fractions decomposition and integrate each fraction. For linear factors , use substitution with u x c =-. For irreducible quadratics, use trig substitution with tan x c d - = ....
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This note was uploaded on 05/06/2011 for the course MATH 256 taught by Professor Buekman during the Spring '11 term at Purdue University-West Lafayette.

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