## Calculus: Early Transcendentals (2nd Edition)

Published by Pearson

# Chapter 4 - Applications of the Derivative - 4.9 Antiderivatives - 4.9 Exercises: 58

#### Answer

$$2{t^5} - 3\ln \left| t \right| + C$$

#### Work Step by Step

\eqalign{ & \int {\frac{{10{t^5} - 3}}{t}} dt \cr & = \int {\left( {\frac{{10{t^5}}}{t} - \frac{3}{t}} \right)dt} \cr & = \int {\left( {10{t^4} - \frac{3}{t}} \right)dt} \cr & {\text{sum rule}} \cr & = \int {10{t^4}dt} - 3\int {\frac{1}{t}dt} \cr & {\text{integrate}} \cr & = 10\left( {\frac{{{t^5}}}{5}} \right) - 3\ln \left| t \right| + C \cr & = 2{t^5} - 3\ln \left| t \right| + C \cr & {\text{check by differentiation}} \cr & {\text{ = }}\frac{d}{{dt}}\left( {2{t^5} - 3\ln \left| t \right| + C} \right) \cr & {\text{ = }}\frac{d}{{dt}}\left( {2{t^5}} \right) - 3\frac{d}{{dt}}\left( {\ln \left| t \right|} \right) + \frac{d}{{dt}}\left( C \right) \cr & {\text{ = }}10t - 3\left( {\frac{1}{t}} \right) + 0 \cr & = 10t - \frac{3}{t} \cr & subtract \cr & = \frac{{10{t^2} - 3}}{t} \cr}

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