Answer
$\log_2 x + \frac{1}{2}\log_2 (x^2+1)$
Work Step by Step
$Expand$ $the$ $logarithmic$ $expression:$
$\log_2 (x\sqrt {x^2+1})$
Use the First Law of Logarithms
$$\log_2 (x\times \sqrt {x^2+1}) = \log_2 x + \log_2 \sqrt {x^2+1}$$
Rewrite the root to exponent form
$$\log_2 x + \log_2 (x^2+1)^{\frac{1}{2}}$$
Use the Third Law of Logarithms for $\log_2 (x^2+1)^{\frac{1}{2}}$
$$\log_2 (x^2+1)^{\frac{1}{2}} = \frac{1}{2}\log_2 (x^2+1)$$
$$\log_2 x + \frac{1}{2}\log_2 (x^2+1)$$