Answer
$$16\ln \left| {x + \sqrt {{x^2} + 16} } \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{16}}{{\sqrt {{x^2} + 16} }}} dx \cr
& or \cr
& = 16\int {\frac{1}{{\sqrt {{x^2} + {{\left( 4 \right)}^2}} }}} dx \cr
& {\text{integrate by tables using the formulas on the apendix D for this book}} \cr
& {\text{using the formula 5}}:\,\,\,\,\,\,\,\int {\frac{1}{{\sqrt {{x^2} + {a^2}} }}dx = \ln \left| {x + \sqrt {{x^2} + {a^2}} } \right|} + C \cr
& {\text{setting }}a = 4{\text{ then}} \cr
& = 16\int {\frac{1}{{\sqrt {{x^2} + 16} }}} dx = 16\left( {\ln \left| {x + \sqrt {{x^2} + {{\left( 4 \right)}^2}} } \right|} \right) + C \cr
& {\text{simplifying}} \cr
& \int {\frac{{16}}{{\sqrt {{x^2} + 16} }}} dx = 16\ln \left| {x + \sqrt {{x^2} + 16} } \right| + C \cr} $$
