## Calculus: Early Transcendentals (2nd Edition)

Published by Pearson

# Chapter 4 - Applications of the Derivative - 4.9 Antiderivatives - 4.9 Exercises - Page 238: 36

#### Answer

$$2{t^6} + \frac{1}{t} + C$$

#### Work Step by Step

\eqalign{ & {\text{split the numerator}} \cr & = \int {\left( {\frac{{12{t^8}}}{{{t^3}}} - \frac{t}{{{t^3}}}} \right)dt} \cr & = \int {\left( {12{t^5} - {t^{ - 2}}} \right)dt} \cr & {\text{use power rule for indefinite integrals}} \cr & = 12\left( {\frac{{{t^6}}}{6}} \right) - \left( {\frac{{{t^{ - 1}}}}{{ - 1}}} \right) + C \cr & = 2{t^6} + \frac{1}{t} + C \cr & {\text{check by differentiation}} \cr & {\text{ = }}\frac{d}{{dt}}\left( {2{t^6} + \frac{1}{t} + C} \right) \cr & = 12{t^5} - \frac{1}{{{t^2}}} + 0 \cr & = \frac{{12{t^7} - 1}}{{{t^2}}} \cr & = \frac{{12{t^8} - t}}{{{t^3}}} \cr}

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