5.5 The Substitution Rule/13

$\int {\frac {dx}{5-3x}}{\text{,}}\quad u=5-3x$

${\begin{aligned}u&=5-3x\\[2ex]du&=-3dx\\[2ex]-{\frac {1}{3}}du&=dx\end{aligned}}$

${\begin{aligned}\int {\frac {dx}{5-3x}}&=-{\frac {1}{3}}\int \left({\frac {1}{u}}\right)du\\[2ex]&=-{\frac {1}{3}}\ln {(u)}du\\[2ex]&=-{\frac {1}{3}}\ln {|5-3x|}+C\end{aligned}}$