Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.1 Exercises - Page 512: 1

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

$ \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 $