## Calculus: Early Transcendentals (2nd Edition)

Published by Pearson

# Chapter 4 - Applications of the Derivative - 4.9 Antiderivatives - 4.9 Exercises: 108

#### Answer

$$x + {\tan ^{ - 1}}x + C$$

#### Work Step by Step

\eqalign{ & \int {\frac{{2 + {x^2}}}{{1 + {x^2}}}dx} \cr & = \int {\frac{{1 + {x^2} + 1}}{{1 + {x^2}}}dx} \cr & {\text{by the long division}} \cr & = \int {\left( {1 + \frac{1}{{1 + {x^2}}}} \right)dx} \cr & {\text{split the integrand}} \cr & = \int {dx} + \int {\frac{1}{{1 + {x^2}}}} dx \cr & {\text{integrate}} \cr & = x + {\tan ^{ - 1}}x + C \cr & {\text{check by differentiation}} \cr & {\text{ = }}\frac{d}{{dx}}\left( {x + {{\tan }^{ - 1}}x + C} \right) \cr & {\text{ = }}\frac{d}{{dx}}\left( x \right) + \frac{d}{{dx}}\left( {{{\tan }^{ - 1}}x} \right) + \frac{d}{{dx}}\left( C \right) \cr & {\text{ = }}1 + \frac{1}{{1 + {x^2}}} + 0 \cr & {\text{simplify}} \cr & {\text{ = }}\frac{{1 + {x^2} + 1}}{{1 + {x^2}}} \cr & add \cr & = \frac{{2 + {x^2}}}{{1 + {x^2}}} \cr}

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