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: 22

Answer

\[\frac{1}{2}\,{\left( {2{y^3} + 3{y^2} + 1} \right)^{\frac{1}{3}}}\]

Work Step by Step

\[\begin{gathered} \int_{}^{} {\frac{{{y^2} + y}}{{\,{{\left( {2{y^3} + 3{y^2} + 1} \right)}^{2/3}}}}dy} \hfill \\ Let\,\,\,u = 2{y^3} + 3{y^2} + 1\,\,,\,\,So\,\,that \hfill \\ du = \,\left( {6{y^2} + 6y} \right)dy \hfill \\ du = 6\,\left( {{y^2} + 6} \right)dy \hfill \\ Then \hfill \\ \frac{1}{6}\int_{}^{} {\frac{{6\,\left( {{y^2} + y} \right)dy}}{{\,{{\left( {2{y^3} + 3{y^2} + 1} \right)}^{2/3}}}} = \frac{1}{6}\int_{}^{} {\frac{{du}}{{{u^{2/3}}}}} } \hfill \\ \frac{1}{6}\int_{}^{} {{u^{ - 2/3}}du} \hfill \\ Power\,\,rule \hfill \\ \frac{1}{6}\,\left( {\frac{{{u^{1/2}}}}{{1/3}}} \right) + C \hfill \\ \frac{1}{2}{u^{\frac{1}{3}}} + C \hfill \\ Substituting\,\,u = 2{y^3} + 3{y^2} + 1\,\,gives \hfill \\ \frac{1}{2}\,{\left( {2{y^3} + 3{y^2} + 1} \right)^{\frac{1}{3}}} \hfill \\ \end{gathered} \]
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