Evaluate the definite integral $\int_{2}^{3} \frac{d x}{x^{2}-1}$.

Let $I = \int_{2}^{3} \frac{d x}{x^{2}-1}$.
Using the standard integral formula $\int \frac{d x}{x^{2}-a^{2}} = \frac{1}{2a} \log \left| \frac{x-a}{x+a} \right| + C$,where $a = 1$,we get:
$F(x) = \int \frac{d x}{x^{2}-1} = \frac{1}{2} \log \left| \frac{x-1}{x+1} \right|$.
By the second fundamental theorem of calculus,$I = F(3) - F(2)$.
$I = \frac{1}{2} \left[ \log \left| \frac{3-1}{3+1} \right| - \log \left| \frac{2-1}{2+1} \right| \right]$.
$I = \frac{1}{2} \left[ \log \left| \frac{2}{4} \right| - \log \left| \frac{1}{3} \right| \right]$.
$I = \frac{1}{2} \left[ \log \left( \frac{1}{2} \right) - \log \left( \frac{1}{3} \right) \right]$.
Using the property $\log a - \log b = \log \left( \frac{a}{b} \right)$:
$I = \frac{1}{2} \log \left( \frac{1/2}{1/3} \right) = \frac{1}{2} \log \left( \frac{3}{2} \right)$.