Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.7 Exercises - Page 380: 20

Answer

$$\frac{1}{2}\ln \left( {{x^2} + 2x + 5} \right) - \frac{3}{2}{\tan ^{ - 1}}\left( {\frac{{x + 1}}{2}} \right) + C$$

Work Step by Step

$$\eqalign{ & \int {\frac{{x - 2}}{{{{\left( {x + 1} \right)}^2} + 4}}} dx \cr & {\text{Let }}u = x + 1,{\text{ }}x = u - 1,{\text{ }}dx = du \cr & {\text{Substitute}} \cr & \int {\frac{{x - 2}}{{{{\left( {x + 1} \right)}^2} + 4}}} dx = \int {\frac{{u - 1 - 2}}{{{u^2} + 4}}} du \cr & = \int {\frac{{u - 3}}{{{u^2} + 4}}} du \cr & {\text{Split the numerator}} \cr & = \int {\frac{u}{{{u^2} + 4}}} du - \int {\frac{3}{{{u^2} + 4}}} du \cr & = \frac{1}{2}\int {\frac{{2u}}{{{u^2} + 4}}} du - 3\int {\frac{1}{{{u^2} + {{\left( 2 \right)}^2}}}} du \cr & {\text{Integrate}} \cr & = \frac{1}{2}\ln \left( {{u^2} + 4} \right) - \frac{3}{2}{\tan ^{ - 1}}\left( {\frac{u}{2}} \right) + C \cr & {\text{Write in terms of }}x \cr & = \frac{1}{2}\ln \left( {{{\left( {x + 1} \right)}^2} + 4} \right) - \frac{3}{2}{\tan ^{ - 1}}\left( {\frac{{x + 1}}{2}} \right) + C \cr & = \frac{1}{2}\ln \left( {{x^2} + 2x + 5} \right) - \frac{3}{2}{\tan ^{ - 1}}\left( {\frac{{x + 1}}{2}} \right) + C \cr} $$
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