5.3 The Fundamental Theorem of Calculus/35

${\begin{aligned}\int _{1}^{9}{\frac {1}{2x}}dx&={\frac {1}{2}}\int _{1}^{9}{\frac {1}{x}}dx\\[2ex]&={\frac {1}{2}}\ln {|x|}{\bigg |}_{1}^{9}\\[2ex]&={\frac {1}{2}}\ln {|9|}-{\frac {1}{2}}\ln {|1|}\\[2ex]&=\ln {|9^{\frac {1}{2}}|}-\ln {|1^{\frac {1}{2}}|}=\ln {3}-0=\ln {3}\\[2ex]\end{aligned}}$