∫ 1 9 1 2 x d x = 1 2 ∫ 1 9 1 x d x = 1 2 ln | x | | 1 9 = 1 2 ln | 9 | − 1 2 ln | 1 | = ln | 9 1 2 | − ln | 1 1 2 | = ln 3 − 0 = ln 3 {\displaystyle {\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\\[2ex]&=\ln {3}\\[2ex]\end{aligned}}}