Answer
$\log_2{x} + \frac{1}{2}\cdot \log_2{(x^2+1)}$
Work Step by Step
Use the rule $\sqrt{a}=a^{\frac{1}{2}}$ to obtain:
$\log_2{(x\sqrt{x^2+1})}=\log_2{(x(x^2+1)^{\frac{1}{2}})}$
RECALL:
(1) $\log_a{(PQ)}= \log_a{P} + \log_a{Q}$
(2) $\log_a{(P^n)}=n \cdot \log_a{P}$
Use rule (1) above to obtain
$\log_2{(x(x^2+1)^{\frac{1}{2}})}=\log_2{x} +\log_2{(x^2+1)^{\frac{1}{2}})}$
Use rule (2) above to obtain:
$\log_2{x} +\log_2{(x^2+1)^{\frac{1}{2}})}=\log_2{x} + \frac{1}{2}\cdot \log_2{(x^2+1)}$
