Answer
$$A = \frac{3}{{80}}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {\left( {16 + {x^2}} \right)^{ - 3/2}},{\text{ }}\left[ {0,3} \right] \cr
& {\text{The area under the curve is given by }} \cr
& A = \int_0^3 {{{\left( {16 + {x^2}} \right)}^{ - 3/2}}} dx \cr
& {\text{Let }}x = 4\tan \theta ,{\text{ }}dx = 4{\sec ^2}\theta d\theta \cr
& {\text{Substituting}} \cr
& \int {{{\left( {16 + {x^2}} \right)}^{ - 3/2}}} dx = \int {{{\left( {16 + {{\left( {4\tan \theta } \right)}^2}} \right)}^{ - 3/2}}\left( {4{{\sec }^2}\theta } \right)} d\theta \cr
& = \int {{{\left( {16 + 16{{\tan }^2}\theta } \right)}^{ - 3/2}}\left( {4{{\sec }^2}\theta } \right)} d\theta \cr
& = \int {{{\left( {16\left[ {1 + {{\tan }^2}\theta } \right]} \right)}^{ - 3/2}}\left( {4{{\sec }^2}\theta } \right)} d\theta \cr
& {\text{Use pythagorean identity }}1 + {\tan ^2}x = {\sec ^2}x \cr
& = \int {{{\left( {16{{\sec }^2}\theta } \right)}^{ - 3/2}}\left( {4{{\sec }^2}\theta } \right)} d\theta \cr
& = \int {{{\left( {16} \right)}^{ - 3/2}}{{\left( {{{\sec }^2}\theta } \right)}^{ - 3/2}}\left( {4{{\sec }^2}\theta } \right)} d\theta \cr
& = \int {\frac{1}{{16}}\left( {{{\sec }^{ - 3}}\theta } \right){{\sec }^2}\theta } d\theta \cr
& = \int {\frac{1}{{16}}{{\left( {\sec \theta } \right)}^{ - 1}}} d\theta \cr
& = \frac{1}{{16}}\int {\cos \theta } d\theta \cr
& = \frac{1}{{16}}\sin \theta + C \cr
& {\text{Where }}\sin \theta = \frac{x}{{\sqrt {{x^2} + 16} }} \cr
& = \frac{x}{{16\sqrt {{x^2} + 16} }} + C \cr
& {\text{Therefore}}{\text{,}} \cr
& A = \int_0^3 {{{\left( {16 + {x^2}} \right)}^{ - 3/2}}} dx \cr
& A = \left[ {\frac{x}{{16\sqrt {{x^2} + 16} }}} \right]_0^3 \cr
& A = \frac{3}{{16\sqrt {{{\left( 3 \right)}^2} + 16} }} - \frac{0}{{2\sqrt {{{\left( 0 \right)}^2} + 16} }} \cr
& A = \frac{3}{{80}} \cr
& \cr
& {\text{Graph}} \cr} $$
