Answer
$\ln x+\frac{1}{2} \ln \left(1+x^{2}\right)$
Work Step by Step
Given: $\quad \ln (x \sqrt{1+x^{2}})$
Use the rule $\quad \ln_a{AB}=\ln_aA+\ln_aB\quad$ to obtain:
\begin{align*}
\ln (x \sqrt{1+x^{2}})&=\ln x+\ln \sqrt{1+x^{2}}\\
\ln (x \sqrt{1+x^{2}})&=\ln x+\ln \left(1+x^{2}\right)^{1/_{2}}
\end{align*}
Using the rule $\ln_ax^{m} = m \ln_a{x}$ gives:
$$\ln (x \sqrt{1+x^{2}})=\ln x+\frac{1}{2} \ln \left(1+x^{2}\right)$$
