## Calculus, 10th Edition (Anton)

Published by Wiley

# Chapter 4 - Integration - 4.2 The Indefinite Integral - Exercises Set 4.2 - Page 279: 12

#### Answer

$$2{x^{1/2}} - \frac{{15{x^{12/5}}}}{{12}} + \frac{1}{9}x + C$$

#### Work Step by Step

\eqalign{ & \int {\left[ {{x^{ - 1/2}} - 3{x^{7/5}} + \frac{1}{9}} \right]dx} \cr & {\text{By the theorem 4}}{\text{.2}}{\text{.3 }}\left( b \right){\text{ and }}\left( c \right) \cr & = \int {{x^{ - 1/2}}dx} - \int {3{x^{7/5}}} dx + \int {\frac{1}{9}} dx \cr & {\text{By the theorem 4}}{\text{.2}}{\text{.3 }}\left( a \right) \cr & = \int {{x^{ - 1/2}}dx} - 3\int {{x^{7/5}}} dx + \frac{1}{9}\int {dx} \cr & {\text{applying the power rule }}\int {{x^n}} dx = \frac{{{x^{n + 1}}}}{{n + 1}} + C \cr & = \frac{{{x^{1/2}}}}{{1/2}} - 3\left( {\frac{{{x^{12/5}}}}{{12/5}}} \right) + \frac{1}{9}x + C \cr & {\text{Simplifying}} \cr & = 2{x^{1/2}} - \frac{{15{x^{12/5}}}}{{12}} + \frac{1}{9}x + C \cr}

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