Calculus: Early Transcendentals (2nd Edition)

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

Chapter 5 - Integration - 5.5 Substitution Rule - 5.5 Exercises: 37

Answer

$$\frac{3}{{112}}{\left( {2x + 1} \right)^{4/3}}\left( {8x - 3} \right) + C$$

Work Step by Step

$$\eqalign{ & \int {x\root 3 \of {2x + 1} dx} \cr & {\text{substitute }}u = 2x + 1,{\text{ }}x = \frac{u}{2} - \frac{1}{2} \cr & {\text{ }}dx = \frac{1}{2}du \cr & \int {x\root 3 \of {2x + 1} dx} = \int {\left( {\frac{u}{2} - \frac{1}{2}} \right)\root 3 \of u \left( {\frac{1}{2}} \right)du} \cr & = \int {\left( {\frac{u}{2} - \frac{1}{2}} \right)\root 3 \of u \left( {\frac{1}{2}} \right)du} \cr & = \frac{1}{4}\int {\left( {{u^{4/3}} - {u^{1/3}}} \right)du} \cr & {\text{find the antiderivative use the power rule}} \cr & = \frac{1}{4}\left( {\frac{3}{7}{u^{7/3}} - \frac{3}{4}{u^{4/3}}} \right) + C \cr & factoring \cr & = \frac{{3{u^{4/3}}}}{{4\left( {28} \right)}}\left( {4u - 21} \right) + C \cr & {\text{ with}}\,\,\,u = 2x + 1 \cr & = \frac{{3{{\left( {2x + 1} \right)}^{4/3}}}}{{112}}\left[ {4\left( {2x + 1} \right) - 7} \right] + C \cr & = \frac{3}{{112}}{\left( {2x + 1} \right)^{4/3}}\left( {8x - 3} \right) + C \cr} $$
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