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The form of the partial fraction decomposition of a rational function is given below. 16 – (x² + 10x) Вх + С + x2 + 4 A %| (x + 2)(x² + 4) x +2 A = B = C = Now evaluate the indefinite integral. [ 16 – (x² + 10x) dx = - (x+ 2)(x² + 4)

The form of the partial fraction decomposition of a rational function is given below. 16 – (x² + 10x) Вх + С + x2 + 4 A %| (x + 2)(x² + 4) x +2 A = B = C = Now evaluate the indefinite integral. [ 16 – (x² + 10x) dx = - (x+ 2)(x² + 4)

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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The form of the partial fraction decomposition of a rational function is given below. 16 – (x² + 10x) Вх + С + x2 +4 A %3D (x + 2)(x² + 4) x +2 A = B = C = Now evaluate the indefinite integral. 16 – (x2 + 10x) -dx = |Tx+ 2)(x² + 4)
Transcribed Image Text:The form of the partial fraction decomposition of a rational function is given below. 16 – (x² + 10x) Вх + С + x2 +4 A %3D (x + 2)(x² + 4) x +2 A = B = C = Now evaluate the indefinite integral. 16 – (x2 + 10x) -dx = |Tx+ 2)(x² + 4)
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