Calculus 10th Edition

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 - 8.4 Exercises - Page 539: 33

Answer

$$\frac{1}{4}\left[ {\frac{{\sqrt 2 }}{2}{{\tan }^{ - 1}}\left( {\frac{x}{{\sqrt 2 }}} \right) + \frac{x}{{{x^2} + 2}}} \right] + C$$

Work Step by Step

$$\eqalign{ & \int {\frac{1}{{4 + 4{x^2} + {x^4}}}} dx \cr & {\text{Factor, recall that }}{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2} \cr & \int {\frac{1}{{4 + 4{x^2} + {x^4}}}} dx = \int {\frac{1}{{{{\left( {2 + {x^2}} \right)}^2}}}} dx \cr & {\text{Refer to the triangle below}} \cr & x = \sqrt 2 \tan \theta ,{\text{ }}dx = \sqrt 2 {\sec ^2}\theta d\theta ,{\text{ }} \cr & {\text{Substituting}} \cr & \int {\frac{1}{{{{\left( {2 + {x^2}} \right)}^2}}}} dx = \int {\frac{1}{{{{\left( {2 + {{\left( {\sqrt 2 \tan \theta } \right)}^2}} \right)}^2}}}} \left( {\sqrt 2 {{\sec }^2}\theta } \right)d\theta \cr & = \int {\frac{1}{{{{\left( {2 + 2{{\tan }^2}\theta } \right)}^2}}}} \left( {\sqrt 2 {{\sec }^2}\theta } \right)d\theta \cr & = \int {\frac{1}{{{2^2}{{\left( {1 + {{\tan }^2}\theta } \right)}^2}}}} \left( {\sqrt 2 {{\sec }^2}\theta } \right)d\theta \cr & = \int {\frac{1}{{{2^2}{{\left( {{{\sec }^2}\theta } \right)}^2}}}} \left( {\sqrt 2 {{\sec }^2}\theta } \right)d\theta \cr & = \frac{{\sqrt 2 }}{4}\int {\frac{1}{{{{\sec }^2}\theta }}} d\theta \cr & = \frac{{\sqrt 2 }}{4}\int {{{\cos }^2}\theta } d\theta \cr & = \frac{{\sqrt 2 }}{4}\int {\left( {\frac{1}{2} + \frac{1}{2}\cos 2\theta } \right)} d\theta \cr & {\text{Integrating}} \cr & = \frac{{\sqrt 2 }}{4}\left( {\frac{1}{2}\theta + \frac{1}{4}\sin 2\theta } \right) + C \cr & = \frac{{\sqrt 2 }}{4}\left( {\frac{1}{2}\theta + \frac{1}{2}\sin \theta \cos \theta } \right) + C \cr & = \frac{{\sqrt 2 }}{8}\theta + \frac{{\sqrt 2 }}{8}\sin \theta \cos \theta + C \cr & {\text{From the triangle, }}\sin \theta = \frac{x}{{\sqrt {{x^2} + 2} }},{\text{ }}\cos \theta = \frac{{\sqrt 2 }}{{\sqrt {{x^2} + 2} }} \cr & = \frac{{\sqrt 2 }}{8}{\tan ^{ - 1}}\left( {\frac{x}{{\sqrt 2 }}} \right) + \frac{{\sqrt 2 }}{8}\left( {\frac{x}{{\sqrt {{x^2} + 2} }}} \right)\left( {\frac{{\sqrt 2 }}{{\sqrt {{x^2} + 2} }}} \right) + C \cr & = \frac{{\sqrt 2 }}{8}{\tan ^{ - 1}}\left( {\frac{x}{{\sqrt 2 }}} \right) + \frac{2}{8}\left( {\frac{x}{{{x^2} + 2}}} \right) + C \cr & = \frac{{\sqrt 2 }}{8}{\tan ^{ - 1}}\left( {\frac{x}{{\sqrt 2 }}} \right) + \frac{x}{{4\left( {{x^2} + 2} \right)}} + C \cr & {\text{Factoring}} \cr & = \frac{1}{4}\left[ {\frac{{\sqrt 2 }}{2}{{\tan }^{ - 1}}\left( {\frac{x}{{\sqrt 2 }}} \right) + \frac{x}{{{x^2} + 2}}} \right] + 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.