## Precalculus: Mathematics for Calculus, 7th Edition

$\frac{1}{2}\ln (x^2-1) - \frac{1}{2}\ln (x^2+1)$
$Expand$ $the$ $logarithmic$ $expression:$ $\ln \sqrt {\frac{x^2-1}{x^2+1}}$ Rewrite the root to exponent form $$\ln (\frac{x^2-1}{x^2+1})^{\frac{1}{2}}$$ Use the Third Law of Logarithms $$\ln (\frac{x^2-1}{x^2+1})^{\frac{1}{2}} = \frac{1}{2}\ln (\frac{x^2-1}{x^2+1})$$ Use the Second Law of Logarithms [Distribute the $\frac{1}{2}$ as well] $$\frac{1}{2}\ln (\frac{x^2-1}{x^2+1}) = \frac{1}{2}\ln (x^2-1) - \frac{1}{2}\ln (x^2+1)$$