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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 UniversityWest Lafayette.
 Spring '11
 BUEKMAN
 Polynomials, Factors, Fractions

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