## Calculus: Early Transcendentals (2nd Edition)

$$- \frac{x}{{36\sqrt {{x^2} - 36} }} + C$$
\eqalign{ & \int {\frac{{dx}}{{{{\left( {{x^2} - 36} \right)}^{3/2}}}}} \cr & {\text{substitute }}x = 6\sec \theta ,{\text{ }}dx = 6\sec \theta \tan \theta d\theta \cr & = \int {\frac{{6\sec \theta \tan \theta d\theta }}{{{{\left( {36{{\sec }^2}\theta - 36} \right)}^{3/2}}}}} \cr & = \int {\frac{{6\sec \theta \tan \theta d\theta }}{{{{\left( {36} \right)}^{3/2}}{{\left( {{{\sec }^2}\theta - 1} \right)}^{3/2}}}}} = \frac{1}{{108}}\int {\frac{{\sec \theta \tan \theta d\theta }}{{{{\left( {{{\tan }^2}\theta } \right)}^{3/2}}}}} \cr & = \frac{1}{{36}}\int {\frac{{\sec \theta d\theta }}{{{{\tan }^2}\theta }}} = \frac{1}{{108}}\int {\left( {\frac{1}{{\cos \theta }}} \right)\left( {\frac{{{{\cos }^2}\theta }}{{{{\sin }^2}\theta }}} \right)} d\theta \cr & = \frac{1}{{36}}\int {\csc \theta \cot \theta } d\theta \cr & {\text{integrating}} \cr & = - \frac{1}{{36}}\csc \theta + C \cr & {\text{substituting }}\csc \theta \cr & = - \frac{1}{{36}}\left( {\frac{x}{{\sqrt {{x^2} - 36} }}} \right) + C \cr & {\text{simplify}} \cr & = - \frac{x}{{36\sqrt {{x^2} - 36} }} + C \cr}