∫ d x 5 − 3 x , u = 5 − 3 x {\displaystyle \int {\frac {dx}{5-3x}}{\text{,}}\quad u=5-3x}
u = 5 − 3 x d u = − 3 d x − 1 3 d u = d x {\displaystyle {\begin{aligned}u&=5-3x\\[2ex]du&=-3dx\\[2ex]-{\frac {1}{3}}du&=dx\end{aligned}}}
∫ d x 5 − 3 x = − 1 3 ∫ ( 1 u ) d u = − 1 3 ln | u | d u = − 1 3 ln | 5 − 3 x | + C {\displaystyle {\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}}}
![{\displaystyle {\begin{aligned}u&=5-3x\\[2ex]du&=-3dx\\[2ex]-{\frac {1}{3}}du&=dx\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/084b592569fb6ffccbc984e9ef712892c68f0b7f)
![{\displaystyle {\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}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/055a40ca0d1c3a8fd13120d0398782e1f212f655)