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The problem 22+3 dæ can be split, like this: S dx +3 f dæ . 1+z² 1+z² What is the result of integrating both "smaller" problems? • S dæ + 3 S ¬ da = 2x · In(1 + a²) +3 · In(1 + æ²) +C 1+z² 1+z² dx + 3 f dx = In(1+x²) +3•arctan ¤ + C dx + 3 [ ,1,dx = x² + ln|x| +3· arctan ¤ +C 1+z²

The problem 22+3 dæ can be split, like this: S dx +3 f dæ . 1+z² 1+z² What is the result of integrating both "smaller" problems? • S dæ + 3 S ¬ da = 2x · In(1 + a²) +3 · In(1 + æ²) +C 1+z² 1+z² dx + 3 f dx = In(1+x²) +3•arctan ¤ + C dx + 3 [ ,1,dx = x² + ln|x| +3· arctan ¤ +C 1+z²

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 problem S dæ can be split, like this: S da + 3 ƒ da . dx . 1+z² 1+z² What is the result of integrating both "smaller" problems? dx + 3 § „dx = 2x · In(1+x²) + 3 . · In(1 + æ²) + C 1+z² 1+z² S dr + 3 S dr = In(1+x²) + 3 • arctan ¤ + C 1+z² 1+z2 S dæ + 3 S „ de = x² + In|¤| +3· arctan ¤ + C dx = x² + ln|¤|+3• arctan x + C 1+z²
Transcribed Image Text:The problem S dæ can be split, like this: S da + 3 ƒ da . dx . 1+z² 1+z² What is the result of integrating both "smaller" problems? dx + 3 § „dx = 2x · In(1+x²) + 3 . · In(1 + æ²) + C 1+z² 1+z² S dr + 3 S dr = In(1+x²) + 3 • arctan ¤ + C 1+z² 1+z2 S dæ + 3 S „ de = x² + In|¤| +3· arctan ¤ + C dx = x² + ln|¤|+3• arctan x + C 1+z²
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