Answer
\[ = \frac{1}{{120}}\ln \left| {\frac{{4x - 15}}{{4x + 15}}} \right| + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\frac{{dx}}{{225 - 16{x^3}}}} \hfill \\
\hfill \\
{\text{rewrite}}\,\,{\text{the}}\,\,{\text{denominator}} \hfill \\
\hfill \\
= \frac{1}{4}\int_{}^{} {\frac{{4dx}}{{{{15}^2} - \,{{\left( {4x} \right)}^2}}}} \hfill \\
\hfill \\
use\,\,\int_{}^{} {\frac{{du}}{{{a^2} - {u^2}}}} = \frac{1}{{2a}}\ln \left| {\frac{{u - a}}{{u + a}}} \right| + C \hfill \\
\hfill \\
with\,\,a = 15{\text{ and }}u = 4x \hfill \\
\hfill \\
therefore \hfill \\
\hfill \\
= \frac{1}{{\,\left( 2 \right)\,\left( {15} \right)\,\left( 4 \right)}}\ln \left| {\frac{{4x - 15}}{{4x + 15}}} \right| + C \hfill \\
\hfill \\
simplify \hfill \\
\hfill \\
= \frac{1}{{120}}\ln \left| {\frac{{4x - 15}}{{4x + 15}}} \right| + C \hfill \\
\end{gathered} \]
