Calculus (3rd Edition)

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

Chapter 8 - Techniques of Integration - 8.3 Trigonometric Substitution - Exercises - Page 410: 39


$$\frac{1}{\sqrt{6}} \ln \left|2 \sqrt{6}\left(\sqrt{6} x+\frac{1}{2 \sqrt{6}}\right)+\sqrt{24\left(6 x^{2}+x+\frac{1}{24}\right)-1}\right|+C$$

Work Step by Step

Given $$\int \frac{d x}{\sqrt{x+6 x^{2}}}$$ Since \begin{align*} x+6 x^{2}&= 6[x^2+\frac{1}{6}x]\\ &=6[(x+1/12)^2-\frac{1}{144}]\\ &=6(x+1/12)^2-\frac{6}{144} \end{align*} Let $$ \sqrt{6}(x+1/12)=\frac{\sqrt{6}}{12}\sec u\ \ \ \to \ \ \ \ \sqrt{6}dx=\frac{\sqrt{6}}{12}\sec u\tan udu$$ Then \begin{align*} \int \frac{d x}{\sqrt{x+6 x^{2}}}&=\int \frac{d x}{\sqrt{6(x+1/12)^2-\frac{6}{144}}}\\ &=\frac{1}{12}\int \frac{\sec u\tan udu}{\sqrt{\frac{6}{144}\sec^2u-\frac{6}{144}}}\\ &=\frac{1}{\sqrt{6}}\int \frac{\sec u\tan udu}{\tan u}\\ & = \frac{1}{\sqrt{6}}\int \sec u du \\ &= \frac{1}{\sqrt{6}}\ln| \sec u+\tan u|+C\\ &= \frac{1}{\sqrt{6}} \ln |2 u \sqrt{6}+\sqrt{24 u^{2}-1}|+C\\ &=\frac{1}{\sqrt{6}} \ln \left|2 \sqrt{6}\left(\sqrt{6} x+\frac{1}{2 \sqrt{6}}\right)+\sqrt{24\left(6 x^{2}+x+\frac{1}{24}\right)-1}\right|+C \end{align*}
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.