## Calculus 10th Edition

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