University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 8 - Practice Exercises - Page 473: 2

Answer

$\dfrac{x^3}{3}\ln x-(\dfrac{x^3}{9})+c$

Work Step by Step

Integration by parts formula suggests that $\int p'(x) q(x)=p(x) q(x)-\int p(x) q'(x)dx$ $\int x^2 \ln x dx=\dfrac{x^3}{3}\ln x-\int (\dfrac{x^3}{3}) (\dfrac{1}{x})dx$ Thus, $\dfrac{x^3}{3}\ln x-\int (\dfrac{x^2}{3})dx=\dfrac{x^3}{3}\ln x-(\dfrac{x^3}{9})+c$
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.