Answer
$\ln{x}+\frac{1}{2}\ln{(1+x^2)}$
Work Step by Step
Recall:
(1) $\sqrt[m]{a}=a^{\frac{1}{m}}$
(2) $\log_a {x^n}=n\cdot \log_a {x}$.
(3) $\log_a{xy}=\log_a{x} +\log_a{y}$
(4) $\log_a{\frac{x}{y}}=\log_a{x} -\log_a{y}$
Use Rule (3) above to obtain:
$\ln{(x\sqrt{1+x^2})}=\ln{x}+\ln{\sqrt{1+x^2}}=\ln{x}+\ln{(1+x^2)^{\frac{1}{2}}}$
Use Rule (2) above to obtain:
$\ln{x}+\ln{(1+x^2)^{\frac{1}{2}}}=\ln{x}+\frac{1}{2}\ln{(1+x^2)}$
