Answer
$log_{2}\frac{x^{2}+6}{x^{2}+1}$
Work Step by Step
$\div$We know that $log_{b}\frac{x}{y}=log_{b}x-log_{b}y$ (where $x$, $y$, and $b$ are positive real numbers and $b\ne1$).
Therefore, $log_{2}(x^{2}+6)-log_{2}(x^{2}+1)=log_{2}\frac{x^{2}+6}{x^{2}+1}$.