Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 5 - The Integral - Chapter Review Exercises - Page 278: 26


$$\int \frac{3x^3-9}{x^2}dx =\frac{3}{2}x^2+\frac{9}{x}+c.$$

Work Step by Step

We have $$\int \frac{3x^3-9}{x^2}dx=\int 3x-9x^{-2}dx=\frac{3}{2}x^2+9x^{-1}+c\\ =\frac{3}{2}x^2+\frac{9}{x}+c.$$
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.