Answer
\[ = \ln \left| {x + \sqrt {{x^2} - 81} } \right| + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\frac{{dx}}{{\sqrt {{x^2} - 81} }}\,\,,\,\,\,\,x > 9} \hfill \\
\hfill \\
the\,\,integral\,\,contains\,\,{a^2} - {x^2}\,\,so \hfill \\
\hfill \\
x = 9\sec \theta \,\,\,\,then\,\,\,\,dx = 9\sec \theta \tan \theta d\theta \hfill \\
and\,\,\sqrt {{x^2} - 81} = 9\tan \theta \hfill \\
\hfill \\
then \hfill \\
\hfill \\
\int_{}^{} {\frac{{dx}}{{\sqrt {{x^2} - 81} }}} = \int_{}^{} {\frac{{9\sec \theta \tan \theta d\theta }}{{9\tan \theta }}} \hfill \\
\hfill \\
simplify \hfill \\
\hfill \\
= \int_{}^{} {\sec \theta d\theta } \hfill \\
\hfill \\
integrate \hfill \\
\hfill \\
= \ln \left| {\sec \theta + \tan \theta } \right| + C \hfill \\
\hfill \\
substitute\,\,back \hfill \\
\hfill \\
= \ln \left| {\frac{x}{9} + \frac{{\sqrt {{x^2} - 81} }}{9}} \right| + C \hfill \\
\hfill \\
simplify \hfill \\
\hfill \\
= \ln \left| {x + \sqrt {{x^2} - 81} } \right| + C \hfill \\
\end{gathered} \]
