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Using tabular integration, the integral S(x3 – 2x – 1)e¬X+1dx is equal to: (x³ + 3x2 + 4x + 3) + C 글 (x3 + 3x2 + 4x + 3) + C the above O the above (-x³ + 3x? – 4x – 3) + C 응 (x3 + 3x2 + 4x + 3) + C

Using tabular integration, the integral S(x3 – 2x – 1)e¬X+1dx is equal to: (x³ + 3x2 + 4x + 3) + C 글 (x3 + 3x2 + 4x + 3) + C the above O the above (-x³ + 3x? – 4x – 3) + C 응 (x3 + 3x2 + 4x + 3) + C

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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Using tabular integration, the integral S(x³ – 2x – 1)e¯X+1dx is equal to: | (x³ + 3x² + 4x + 3) + C (x³ + 3x2 + 4x + 3) + C the above the above (-x³ + 3x² – 4x – 3) + C (x³ + 3x? + 4x + 3) + C the above the above none of them
Transcribed Image Text:Using tabular integration, the integral S(x³ – 2x – 1)e¯X+1dx is equal to: | (x³ + 3x² + 4x + 3) + C (x³ + 3x2 + 4x + 3) + C the above the above (-x³ + 3x² – 4x – 3) + C (x³ + 3x? + 4x + 3) + C the above the above none of them
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