Answer
\[ = \frac{1}{{15\,{{\left( {3 - 5x} \right)}^3}}} + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\frac{{dx}}{{\,{{\left( {3 - 5x} \right)}^4}}}} \hfill \\
\hfill \\
substituting\, \hfill \\
\hfill \\
3 - 5x = u\,\,\,\,and\,\,\, - 5x = du\,\,or\,\,\,dx = - \frac{{du}}{5} \hfill \\
\hfill \\
- \frac{1}{5}\int_{}^{} {\frac{{du}}{{{u^4}}}} = - \frac{1}{5}\int {{u^{ - 4}}du} \hfill \\
\hfill \\
{\text{use }}\int {{x^n}dx = \frac{{{x^{n + 1}}}}{{n + 1}} + C} \hfill \\
\hfill \\
= - \frac{1}{5}\,\left( { - \frac{1}{{3{u^3}}}} \right) + C = \frac{1}{{15{u^3}}} + C \hfill \\
\hfill \\
substituting\,back\,u = \,\left( {3 - 5x} \right) \hfill \\
\hfill \\
= \frac{1}{{15\,{{\left( {3 - 5x} \right)}^3}}} + C \hfill \\
\end{gathered} \]
