Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 7 - Review - Exercises - Page 538: 14


$\displaystyle \frac{1}{2}x^{2}-2x+6\ln|x+2|+C$

Work Step by Step

$\displaystyle \frac{x^{2}+2}{x+2}=\frac{x(x+2)-2x+2}{x+2}=x+\frac{-2x+2}{x+2}$ $=x+\displaystyle \frac{-2(x+2)+6}{x+2}=x-2+\frac{6}{x+2}$ $\displaystyle \int\frac{x^{2}+2}{x+2}dx=\int\left(x-2+\frac{6}{x+2}\right)dx $ $=\displaystyle \frac{1}{2}x^{2}-2x+6\ln|x+2|+C$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.