Answer
\[ = \frac{{\ln \left| {10x - 3} \right|}}{{10}} + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\frac{1}{{10x - 3}}} \,dx \hfill \\
\hfill \\
set\,\,the\,\,substitution \hfill \\
\hfill \\
u = 10x - 3\,\,\,then\,\,\,du = 10xdx \hfill \\
\hfill \\
therefore \hfill \\
\hfill \\
\int_{}^{} {\frac{1}{{10x - 3}}\,dx} = \frac{1}{{10}} \cdot \int_{}^{} {\frac{{10}}{{10x - 3}}} \,dx \hfill \\
\hfill \\
\frac{1}{{10}} \cdot \int_{}^{} {\frac{1}{u}\,du} \hfill \\
\hfill \\
integrate\,\, \hfill \\
\hfill \\
= \frac{{\ln \left| u \right|}}{{10}} + C \hfill \\
\hfill \\
replace\,\,u\,\,with\,\,\,u = 10x - 3 \hfill \\
\hfill \\
= \frac{{\ln \left| {10x - 3} \right|}}{{10}} + C \hfill \\
\end{gathered} \]
