Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.1 Exercises: 20

Answer

$2x^2+\frac 1 {2x+3}+C$

Work Step by Step

$\int 4x -\frac 2 {(2x+3)^2}dx$ $4\int xdx -2\int \frac 1 {(2x+3)^2}dx$ $2x^2-\frac {(2x+3)^-1}{-1}+C$ $2x^2+\frac 1 {2x+3}+C$
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