Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 7 - Integration - 7.2 Substitution - 7.2 Exercises: 19

Answer

\[\frac{1}{4}\ln \,\left( {{x^4} + 4{x^2} + 7} \right) + C\]

Work Step by Step

\[\begin{gathered} \int_{}^{} {\frac{{{x^3} + 2x}}{{{x^4} + 4{x^2} + 7}}dx} \hfill \\ Let\,\,u = {x^4} + 4{x^2} + 7 \hfill \\ So\,\,that \hfill \\ du = \,\left( {4{x^3} + 8x} \right)dx \hfill \\ du = 4\,\left( {{x^3} + 2x} \right)dx \hfill \\ \int_{}^{} {\frac{{{x^3} + 2x}}{{{x^4} + 4{x^2} + 7}}dx = \frac{1}{4}\int_{}^{} {\frac{{4\,\left( {{x^3} + 2x} \right)dx}}{{{x^4} + 4{x^2} + 7}}} } \hfill \\ Integrating \hfill \\ \frac{1}{4}\int_{}^{} {\frac{{du}}{u} = \frac{1}{4}\ln \left| u \right| + C} \hfill \\ Substituting\,\,u = {x^4} + 4{x^2} + 7\,\,for\,\,u\,\,gives \hfill \\ \frac{1}{4}\ln \,\left( {{x^4} + 4{x^2} + 7} \right) + C \hfill \\ \end{gathered} \]
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