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: 34

Answer

$$ - \frac{1}{{y + 1}} + \frac{1}{{{{\left( {y + 1} \right)}^2}}} - \frac{1}{{3{{\left( {y + 1} \right)}^3}}} + C$$

Work Step by Step

$$\eqalign{ & \int {\frac{{{y^2}}}{{{{\left( {y + 1} \right)}^4}}}dy} \cr & {\text{substitute }}u = y + 1,{\text{ }}y = u - 1 \cr & {\text{ }}dy = du \cr & \int {\frac{{{y^2}}}{{{{\left( {y + 1} \right)}^4}}}dy} = \int {\frac{{{{\left( {u - 1} \right)}^2}}}{{{u^4}}}du} \cr & = \int {\frac{{{u^2} - 2u + 1}}{{{u^4}}}du} \cr & = \int {\left( {\frac{{{u^2}}}{{{u^4}}} - \frac{{2u}}{{{u^4}}} + \frac{1}{{{u^4}}}} \right)du} \cr & = \int {\left( {\frac{1}{{{u^2}}} - \frac{2}{{{u^3}}} + \frac{1}{{{u^4}}}} \right)du} \cr & = \int {\left( {{u^{ - 2}} - 2{u^{ - 3}} + {u^{ - 4}}} \right)du} \cr & {\text{find the antiderivative use the power rule}} \cr & = \frac{{{u^{ - 1}}}}{{ - 1}} - 2\left( {\frac{{{u^{ - 2}}}}{{ - 2}}} \right) + \frac{{{u^{ - 3}}}}{{ - 3}} + C \cr & = - \frac{1}{u} + \frac{1}{{{u^2}}} - \frac{1}{{3{u^3}}} + C \cr & {\text{ with}}\,\,\,u = y + 1 \cr & = - \frac{1}{{y + 1}} + \frac{1}{{{{\left( {y + 1} \right)}^2}}} - \frac{1}{{3{{\left( {y + 1} \right)}^3}}} + C \cr} $$
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