Calculus: Early Transcendentals (2nd Edition)

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

Chapter 7 - Integration Techniques - 7.8 Improper Integrals - 7.8 Exercises: 20

Answer

\[ = \frac{\pi }{2}\]

Work Step by Step

\[\begin{gathered} \int_{ - \infty }^\infty {\frac{{dx}}{{{x^2} + 2x + 5}}} = \int_{}^{} {\frac{{dx}}{{{x^2} + 2x + 1 + 4}}} \hfill \\ \hfill \\ = \int_{}^{} {\frac{{dx}}{{\,{{\left( {x + 1} \right)}^2} + 4}}} = \frac{1}{2}{\tan ^{ - 1}}\frac{{x + 1}}{2} + C \hfill \\ \hfill \\ If\,\,f\,\left( x \right)\,\,is\,\,continuous\,\,on\,\,\left( { - \infty ,\infty } \right]\,\,then \hfill \\ \hfill \\ \int_{ - \infty }^\infty {f\,\left( x \right)} \,dx = \,\,\,\,\,\mathop {\,\lim }\limits_{a \to - \infty } \int_a^c {f\,\left( x \right)dx} + \,\mathop {\,\lim }\limits_{b \to \infty } \int_c^b {f\,\left( x \right)dx} \hfill \\ \hfill \\ use\,\,Definition\,\,of\,\,improper\,\,integral\,\, \hfill \\ \hfill \\ \int_{ - \infty }^\infty {\frac{{dx}}{{{x^2} + 2x + 5}}} = \mathop {\lim }\limits_{a \to - \infty } \int_a^c {\frac{{dx}}{{{x^2} + 2x + 5}}\,dx} + \mathop {\lim }\limits_{b \to \infty } \int_c^b {\frac{{dx}}{{{x^2} + 2x + 5}}} dx \hfill \\ \hfill \\ \operatorname{int} egrate \hfill \\ \hfill \\ = \mathop {\lim }\limits_{a \to - \infty } \,\left( {\frac{1}{2}{{\tan }^{ - 1}}\frac{{x + 1}}{2}} \right)_a^0 + \mathop {\lim }\limits_{b \to \infty } \left( {\frac{1}{2}{{\tan }^{ - 1}}\frac{{x + 1}}{2}} \right)_0^b \hfill \\ \hfill \\ use\,\,the\,\,ftc\,\, \hfill \\ = \mathop {\lim }\limits_{a \to - \infty } \,\left( {\frac{1}{2}{{\tan }^{ - 1}}\frac{{0 + 1}}{2} - \frac{1}{2}{{\tan }^{ - 1}}\frac{{b + 1}}{2}} \right) + \mathop {\lim }\limits_{b \to \infty } \,\left( {\frac{1}{2}{{\tan }^{ - 1}}\frac{{a + 1}}{2} - \frac{1}{2}{{\tan }^{ - 1}}\frac{{0 + 1}}{2}} \right) \hfill \\ \hfill \\ = \mathop {\lim }\limits_{a \to - \infty } \,\left( {\frac{1}{2}{{\tan }^{ - 1}}\frac{{b + 1}}{2}} \right)\, + \,\mathop {\lim }\limits_{b \to \infty } \,\left( {\frac{1}{2}{{\tan }^{ - 1}}\frac{{a + 1}}{2}} \right) \hfill \\ evaluate\,\,the\,\,\lim it \hfill \\ \hfill \\ = \frac{1}{2}\mathop {\lim }\limits_{a \to - \infty } \,\left( {\frac{\pi }{2} - \,\left( { - \frac{\pi }{2}} \right)} \right) = \frac{\pi }{2} \hfill \\ \end{gathered} \]
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.