Calculus with Applications (10th Edition)

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

Chapter 7 - Integration - Chapter Review - Review Exercises - Page 418: 35

Answer

$$ - \frac{1}{{9{{\left( {{x^3} + 5} \right)}^3}}} + C$$

Work Step by Step

$$\eqalign{ & \int {\frac{{{x^2}dx}}{{{{\left( {{x^3} + 5} \right)}^4}}}} \cr & {\text{set }}u = {x^3} + 5{\text{ then }}\frac{{du}}{{dx}} = 3{x^2},\,\,\,\,\,\,\,\,{x^2}dx = \frac{1}{3}du \cr & {\text{write the integrand in terms of }}u \cr & \int {\frac{{{x^2}dx}}{{{{\left( {{x^3} + 5} \right)}^4}}}} = \int {\frac{{1/3}}{{{u^4}}}} du \cr & = \frac{1}{3}\int {\frac{1}{{{u^4}}}} du \cr & {\text{rewrite with negative exponent using }}\frac{1}{{{u^n}}} = {u^{ - n}} \cr & = \frac{1}{3}\int {{u^{ - 4}}} du \cr & {\text{integrate by using the power rule }}\int {{u^n}} du = \frac{{{u^{n + 1}}}}{{n + 1}} + C \cr & = \frac{1}{3}\left( {\frac{{{u^{ - 4 + 1}}}}{{ - 4 + 1}}} \right) + C \cr & = \frac{1}{3}\left( {\frac{{{u^{ - 3}}}}{{ - 3}}} \right) + C \cr & = - \frac{1}{9}{u^{ - 3}} + C \cr & = - \frac{1}{{9{u^3}}} + C \cr & {\text{replace }}{x^3} + 5{\text{ for }}u \cr & = - \frac{1}{{9{{\left( {{x^3} + 5} \right)}^3}}} + C \cr} $$
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