Answer
$$A = \frac{{3\pi }}{8}$$
Work Step by Step
$$\eqalign{
& y = \frac{6}{{16 + {x^2}}} \cr
& {\text{The area of the region is given by}} \cr
& A = \int_0^4 {\frac{6}{{16 + {x^2}}}} dx \cr
& A = 6\int_0^4 {\frac{1}{{{{\left( 4 \right)}^2} + {x^2}}}} dx \cr
& {\text{Integrate using the basic integration formula}} \cr
& A = 6\left( {\frac{1}{4}{{\tan }^{ - 1}}\left( {\frac{x}{4}} \right)} \right)_0^4 \cr
& A = \frac{3}{2}\left( {{{\tan }^{ - 1}}\left( {\frac{x}{4}} \right)} \right)_0^4 \cr
& {\text{Evaluate and simplify}} \cr
& A = \frac{3}{2}\left( {{{\tan }^{ - 1}}\left( {\frac{4}{4}} \right) - {{\tan }^{ - 1}}\left( {\frac{0}{4}} \right)} \right) \cr
& {\text{Simplifying}} \cr
& A = \frac{3}{2}\left( {\frac{\pi }{4}} \right) \cr
& A = \frac{{3\pi }}{8} \cr} $$
