Calculus (3rd Edition)

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

Chapter 8 - Techniques of Integration - Chapter Review Exercises - Page 460: 45


$$\frac{1}{3}\tan^{-1}\left(\frac{x+4}{3}\right)+C $$

Work Step by Step

Given $$ \int \frac{d x}{x^{2}+8 x+25}$$ Since $$x^2+8x+25= (x+4)^2+9 $$ Then \begin{aligned} \int \frac{d x}{x^{2}+8 x+25} &=\int \frac{d x}{(x+4)^{2}+3^{2}}\\ &=\frac{1}{3}\tan^{-1}\left(\frac{x+4}{3}\right)+C \end{aligned}
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.