Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 5 - Integration - 5.5 Substitution Rule - 5.5 Exercises: 30

Answer

$$\frac{3}{5}{\tan ^{ - 1}}5y + C$$

Work Step by Step

$$\eqalign{ & \int {\frac{3}{{1 + 25{y^2}}}dy} \cr & {\text{substitute }}u = 5y,{\text{ }}du = 5dy{\text{ and }}dy = \frac{{du}}{5} \cr & \int {\frac{3}{{1 + 25{y^2}}}dy} = \int {\frac{3}{{1 + {u^2}}}\left( {\frac{{du}}{5}} \right)} \cr & {\text{take out the constant}} \cr & = \frac{3}{5}\int {\frac{{du}}{{1 + {u^2}}}} \cr & {\text{find the antiderivative}} \cr & = \frac{3}{5}{\tan ^{ - 1}}u + C \cr & {\text{ with}}\,\,\,u = 5y \cr & = \frac{3}{5}{\tan ^{ - 1}}5y + C \cr} $$
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.