Calculus: Early Transcendentals (2nd Edition)

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

Chapter 7 - Integration Techniques - 7.1 Basic Approaches - 7.1 Exercises - Page 515: 57

Answer

\[\begin{align} & \text{a}\text{. }\frac{1}{2}{{x}^{2}}-x+\ln \left| x+1 \right|+C \\ & \text{b}\text{. }\frac{1}{2}{{x}^{2}}-x+\ln \left| x+1 \right|+C \\ & \text{c}\text{.The antiderivatives differ by a constant}\text{.} \\ \end{align}\]

Work Step by Step

\[\begin{align} & \text{a}\text{. }\int{\frac{{{x}^{2}}}{x+1}}dx \\ & \text{Let }u=x+1,\text{ }x=u-1,\text{ }dx=du \\ & \int{\frac{{{x}^{2}}}{x+1}}dx=\int{\frac{{{\left( u-1 \right)}^{2}}}{u}}du \\ & \text{ }=\int{\frac{{{u}^{2}}-2u+1}{u}}du \\ & \text{ }=\int{\left( u-2+\frac{1}{u} \right)}du \\ & \text{Integrating} \\ & =\frac{1}{2}{{u}^{2}}-2u+\ln \left| u \right|+C \\ & \text{Back-substitute }u=x+1 \\ & =\frac{1}{2}{{\left( x+1 \right)}^{2}}-2\left( x+1 \right)+\ln \left| x+1 \right|+C \\ & \text{Simplify and combine constants} \\ & =\frac{1}{2}\left( {{x}^{2}}+2x+1 \right)-2x-2+\ln \left| x+1 \right|+C \\ & =\frac{1}{2}{{x}^{2}}+x+\frac{1}{2}-2x-2+\ln \left| x+1 \right|+C \\ & =\frac{1}{2}{{x}^{2}}-x+\ln \left| x+1 \right|+{{C}_{1}} \\ & \\ & \text{b}\text{. }\int{\frac{{{x}^{2}}}{x+1}}dx \\ & \text{By the long division} \\ & \int{\frac{{{x}^{2}}}{x+1}}dx=\int{\left( x-1+\frac{1}{x+1} \right)}dx \\ & \text{Integrating} \\ & =\frac{1}{2}{{x}^{2}}-x+\ln \left| x+1 \right|+C \\ & \\ & \text{c}\text{.The antiderivatives differ by a constant}\text{.} \\ \end{align}\]
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