## Calculus: Early Transcendentals (2nd Edition)

Published by Pearson

# Chapter 4 - Applications of the Derivative - 4.9 Antiderivatives - 4.9 Exercises - Page 238: 50

#### Answer

$$\frac{3}{2}{\tan ^{ - 1}}\frac{v}{2} + C$$

#### Work Step by Step

\eqalign{ & \int {\frac{3}{{4 + {v^2}}}} dv \cr & {\text{take out the constant}} \cr & = 3\int {\frac{1}{{4 + {v^2}}}} dv \cr & {\text{from the table 4}}{\text{.10 }}\int {\frac{{dx}}{{{a^2} + {x^2}}}} = \frac{1}{a}{\tan ^{ - 1}}\frac{x}{a} + C \cr & {\text{letting }}a = 2,{\text{ }}v = x \cr & = 3\left( {\frac{1}{2}{{\tan }^{ - 1}}\frac{v}{2}} \right) + C \cr & = \frac{3}{2}{\tan ^{ - 1}}\frac{v}{2} + C \cr & {\text{check by differentiation}} \cr & {\text{ = }}\frac{d}{{dv}}\left( {\frac{3}{2}{{\tan }^{ - 1}}\frac{v}{2} + C} \right) \cr & {\text{ = }}\frac{{\text{3}}}{2}\frac{d}{{dv}}\left( {{{\tan }^{ - 1}}\frac{v}{2}} \right) + \frac{d}{{dv}}\left( C \right) \cr & {\text{ = }}\frac{{\text{3}}}{2}\left( {\frac{2}{{{{\left( 2 \right)}^2} + {v^2}}}} \right) + 0 \cr & = \frac{3}{{4 + {v^2}}} \cr}

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