For the part highlighted in green, I'm lost where these parts came from. 20xABx +C8.5.55 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%3Dso B = -2. The original integral is thus/(-) -/ (은2x - 10(2x+4)-14dxdx =x-1 x2 +4x +5x - 1x² + 4x + 5(2x+4)2+ 4x +5 (x+ 2)2 + 114dxx - 1(x – 1)²= Inx2 + 4x + 5+ 14 tan-(x + 2) + C.%3D

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Answered: 2x- 10 dx = (2x+4)-14 dx - 1 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

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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