Using the substitution: u = x + 9. Re-write the indefinite integral then evaluate in term x2 dr = / Vx + 9 Note: answer should be in terms of u only Part 2. Back substituting in the antiderivative you found in Part 1. above we have x2 dx = Vx + 9
Using the substitution: u = x + 9. Re-write the indefinite integral then evaluate in term x2 dr = / Vx + 9 Note: answer should be in terms of u only Part 2. Back substituting in the antiderivative you found in Part 1. above we have x2 dx = Vx + 9
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter2: Equations And Inequalities
Section2.3: Quadratic Equations
Problem 51E
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