Answer
The integral is improper.
Work Step by Step
The integral has bounds of $[1,\infty]$, therefore, it is improper. In addition, the integrand $\ln(x^2)$ approaches a vertical asymptote as $x\to0^+$. This also makes the integral improper.
Calculus 10th Edition
by Larson, Ron; Edwards, Bruce H.
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 - 8.8 Exercises - Page 575: 4
Update this answer!
You can help us out by revising, improving and updating this answer.
Update this answerAfter 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.
