Answer
\[ = - \frac{1}{{18\,{{\left( {9x - 2} \right)}^2}}} + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\,{{\left( {9x - 2} \right)}^{ - 3}}dx} \hfill \\
\hfill \\
substituting\,\,\,\left( {9x - 2} \right) = u\,\,\,\,and\,\,\,\,dx = \frac{{du}}{9} \hfill \\
\hfill \\
= \frac{1}{9}\int_{}^{} {{u^{ - 3}}\,du} \hfill \\
\hfill \\
{\text{use }}\int {{x^n}dx = \frac{{{x^{n + 1}}}}{{n + 1}} + C} \hfill \\
\hfill \\
= \frac{1}{9}\,\left( {\frac{{ - 1}}{{2{u^2}}}} \right) + C = - \frac{1}{{18{u^2}}} + C \hfill \\
\hfill \\
substituting\,\,back\,\,u = \,\left( {9x - 2} \right) \hfill \\
\hfill \\
= - \frac{1}{{18\,{{\left( {9x - 2} \right)}^2}}} + C \hfill \\
\end{gathered} \]
