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: 18

Answer

\[ - 2\ln \,\left( {{x^2} + 3} \right) + C\]

Work Step by Step

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