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Tutorial Exercise Use partial fractions to find the indefinite integral. 16x1 dx Step 1 Rewrite the polynomial in the denominator as 16x¹ - 1 = (4x²)² - 12 = (4x²-1)(C -((2x)²-13)(C = (2x - 1)(2x + 1)( x² + 1) your skip a part of Jx²+1)

Tutorial Exercise Use partial fractions to find the indefinite integral. 16x1 dx Step 1 Rewrite the polynomial in the denominator as 16x¹ - 1 = (4x²)² - 12 = (4x²-1)(C -((2x)²-13)(C = (2x - 1)(2x + 1)( x² + 1) your skip a part of Jx²+1)

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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Tutorial Exercise Use partial fractions to find the indefinite integral. 16x²-1² dx Step 1 Rewrite the polynomial in the denominator as 16x¹ - 1 = (4x2)2 - 1² = (4x²-1)(C =((2x)²-1²)(C = (2x - 1)(2x + 1) (1 x²+1). you
Transcribed Image Text:Tutorial Exercise Use partial fractions to find the indefinite integral. 16x²-1² dx Step 1 Rewrite the polynomial in the denominator as 16x¹ - 1 = (4x2)2 - 1² = (4x²-1)(C =((2x)²-1²)(C = (2x - 1)(2x + 1) (1 x²+1). you
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