Calculus: Early Transcendentals (2nd Edition)

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

Chapter 7 - Integration Techniques - 7.5 Partial Fractions - 7.5 Exercises - Page 549: 48

Answer

\[ = \ln \,\left( {{x^2} + 4} \right) + \frac{1}{2}{\tan ^{ - 1}}\,\left( {\frac{x}{2}} \right) + C\]

Work Step by Step

\[\begin{gathered} \int_{}^{} {\frac{{2x + 1}}{{{x^2} + 4}}} \hfill \\ \hfill \\ split\,\,the\,\,numerator \hfill \\ \hfill \\ \frac{{2x + 1}}{{{x^2} + 4}} = \frac{{2x}}{{{x^2} + 4}} + \frac{1}{{{x^2} + 4}} \hfill \\ \hfill \\ {\text{Therefore}}{\text{,}} \hfill \\ \hfill \\ \int_{}^{} {\frac{{2x + 1}}{{{x^2} + 4}}} = \int_{}^{} {\,\left( {\frac{{2x}}{{{x^2} + 4}} + \frac{1}{{{x^2} + 4}}} \right)dx} \hfill \\ \hfill \\ {\text{integrating}} \hfill \\ \hfill \\ = \ln \,\left( {{x^2} + 4} \right) + \frac{1}{2}{\tan ^{ - 1}}\,\left( {\frac{x}{2}} \right) + C \hfill \\ \hfill \\ \end{gathered} \]
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