Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 4 - Applications of the Derivative - 4.9 Antiderivatives - 4.9 Exercises: 107

Answer

$$\ln \left| x \right| + 2{x^{1/2}} + C$$

Work Step by Step

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