Answer
$$\frac{{16{x^2} - 64}}{{x\sqrt {16 - {x^2}} }} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{x^3}}}{{{{\left( {{x^2} - 16} \right)}^{3/2}}}}} dx,\,\,\,\,x < - 4 \cr
& {\text{The integrand contains the form }}{x^2} - {a^2} \cr
& {x^2} - 16 \to a = 4 \cr
& {\text{Use the change of variable }}x = a\sec \theta \cr
& x = 4\sec \theta ,\,\,\,dx = 4\sec \theta \tan \theta d\theta \cr
& \cr
& {\text{Substituting}} \cr
& {\text{ = }}\int {\frac{{{{\left( {4\sec \theta } \right)}^3}}}{{{{\left( {16{{\sec }^2}\theta - 16} \right)}^{3/2}}}}} \left( {4\sec \theta \tan \theta } \right)d\theta \cr
& {\text{ = }}\int {\frac{{64{{\sec }^3}\theta }}{{64{{\left( {{{\sec }^2}\theta - 1} \right)}^{3/2}}}}} \left( {4\sec \theta \tan \theta } \right)d\theta \cr
& {\text{ = }}\int {\frac{{{{\sec }^3}\theta }}{{{{\left( {{{\tan }^2}\theta } \right)}^{3/2}}}}} \left( {4\sec \theta \tan \theta } \right)d\theta \cr
& {\text{ = }}\int {\frac{{{{\sec }^3}\theta }}{{{{\tan }^3}\theta }}} \left( {4\sec \theta \tan \theta } \right)d\theta \cr
& = 4\int {\frac{{{{\sec }^4}\theta }}{{{{\tan }^2}\theta }}} d\theta \cr
& = 4\int {{{\left( {\tan \theta } \right)}^{ - 2}}{{\sec }^4}\theta } d\theta \cr
& = 4\int {{{\left( {\tan \theta } \right)}^{ - 2}}{{\sec }^2}\theta {{\sec }^2}\theta } d\theta \cr
& = 4\int {{{\left( {\tan \theta } \right)}^{ - 2}}\left( {{{\tan }^2}\theta + 1} \right){{\sec }^2}\theta } d\theta \cr
& = 4\int {\left( {1 + {{\tan }^{ - 2}}\theta } \right){{\sec }^2}\theta } d\theta \cr
& {\text{Integrating}} \cr
& = 4\tan \theta - 4{\left( {\tan \theta } \right)^{ - 1}} + C \cr
& = 4\tan \theta - 4\cot \theta + C \cr
& {\text{Write in terms of }}x \cr
& = 4\left( {\frac{x}{{\sqrt {16 - {x^2}} }}} \right) - 4\left( {\frac{{\sqrt {16 - {x^2}} }}{x}} \right) + C \cr
& = \frac{{4x}}{{\sqrt {16 - {x^2}} }} - \frac{{4\sqrt {16 - {x^2}} }}{x} + C \cr
& = \frac{{4{x^2} - 4\left( {16 - {x^2}} \right)}}{{x\sqrt {16 - {x^2}} }} + C \cr
& = \frac{{16{x^2} - 64}}{{x\sqrt {16 - {x^2}} }} + C \cr} $$
