Question:
$\int_{2}^{3} \frac{1}{x} d x$
Solution:
Let $I=\int_{2}^{3} \frac{1}{x} d x$
$\int \frac{1}{x} d x=\log |x|=\mathrm{F}(x)$
By second fundamental theorem of calculus, we obtain
$\begin{aligned} I &=\mathrm{F}(3)-\mathrm{F}(2) \\ &=\log |3|-\log |2|=\log \frac{3}{2} \end{aligned}$