Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.1 Exercises: 1

Answer

(b) $\sqrt{x^2+1}+C$

Work Step by Step

$ \frac{dy}{dx} = \frac{x}{ \sqrt{x^2+1} } $ $ \int{ \frac{x}{ \sqrt{x^2+1} } dx} $ $ \int{ \frac{x}{ \sqrt{u} } \frac{du}{2x} } $ Substitution: $ u=x^2+1 $ $ \frac{1}{2} \int{\frac{1}{ \sqrt{u} } du } $ $ \frac{1}{2} (2 \sqrt{u} + C) $ $ \sqrt{u} + C $ $ \sqrt{x^2+1} + C $
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