Calculus: Early Transcendentals 8th Edition

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

Chapter 5 - Review - Exercises - Page 422: 25

Answer

$\int (\frac{1-x}{x})^2~dx = -\frac{1}{x}-2~ln~\vert x \vert+x+C$

Work Step by Step

$\int (\frac{1-x}{x})^2~dx$ $=\int (\frac{1-2x+x^2}{x^2})~dx$ $=\int (\frac{1}{x^2}-\frac{2x}{x^2}+\frac{x^2}{x^2})~dx$ $=\int (\frac{1}{x^2}-\frac{2}{x}+1)~dx$ $= -\frac{1}{x}-2~ln~\vert x \vert+x+C$
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.