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2x- 10 dx = (2x+4)-14 dx - 1 x2 + 4x +5 x2 +4x+5 (2x+4) 7? + 4x +5 (x+ 2)2+1 14 dx 1 (*- 1)2 = In + 14 tan- (x + 2) + C. x2+4x +5

2x- 10 dx = (2x+4)-14 dx - 1 x2 + 4x +5 x2 +4x+5 (2x+4) 7? + 4x +5 (x+ 2)2+1 14 dx 1 (*- 1)2 = In + 14 tan- (x + 2) + C. x2+4x +5

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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20x A Bx +C 8.5.55 If we writę (x- 1)(x² + 4x +5) then 20x = A(x² +4x+5)+(Br+C)(x-1). x2+4x+5' Letting r = 1 yields A = 2. Letting æ =0 yields 0 = 10--C, so C = 10. Letting x = 2 yields 40 = 34+2B+10, x-1 %3D so B = -2. The original integral is thus /(-) -/ (은 2x - 10 (2x+4)-14 dx dx = x-1 x2 +4x +5 x - 1 x² + 4x + 5 (2x+4) 2+ 4x +5 (x+ 2)2 + 1 14 dx x - 1 (x – 1)² = In x2 + 4x + 5 + 14 tan-(x + 2) + C. %3D
Transcribed Image Text:20x A Bx +C 8.5.55 If we writę (x- 1)(x² + 4x +5) then 20x = A(x² +4x+5)+(Br+C)(x-1). x2+4x+5' Letting r = 1 yields A = 2. Letting æ =0 yields 0 = 10--C, so C = 10. Letting x = 2 yields 40 = 34+2B+10, x-1 %3D so B = -2. The original integral is thus /(-) -/ (은 2x - 10 (2x+4)-14 dx dx = x-1 x2 +4x +5 x - 1 x² + 4x + 5 (2x+4) 2+ 4x +5 (x+ 2)2 + 1 14 dx x - 1 (x – 1)² = In x2 + 4x + 5 + 14 tan-(x + 2) + C. %3D
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