Answer
\[ = - \frac{1}{4}\frac{1}{{{{\tan }^4}x}} + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\frac{{{{\sec }^2}x}}{{{{\tan }^5}x}}} dx \hfill \\
\hfill \\
use\,\,\tan \,x = u\,\,\,\,\,then\,\,\,{\sec ^2}xdx = du \hfill \\
\hfill \\
= \int_{}^{} {\frac{{du}}{{{u^5}}}} = \hfill \\
\hfill \\
rewrite \hfill \\
\hfill \\
\int_{}^{} {{u^{ - 5}}du} \hfill \\
\hfill \\
integrate \hfill \\
\hfill \\
= \frac{{{u^{ - 4}}}}{{ - 4}} + C \hfill \\
\hfill \\
substituting\,\,back\,\,u\, = \,\tan x \hfill \\
\hfill \\
= - \frac{1}{4}\frac{1}{{{{\tan }^4}x}} + C \hfill \\
\hfill \\
\hfill \\
\end{gathered} \]
