Answer
$$\frac{1}{8}\ln \left| {\frac{{x + 4}}{{x - 4}}} \right| + C\,\,\,$$
Work Step by Step
$$\eqalign{
& \int {\frac{1}{{16 - {x^2}}}} dx \cr
& {\text{Use the Endpaper Integral Table to evaluate the integral}} \cr
& {\text{Rewrite the integrand}} \cr
& \int {\frac{1}{{16 - {x^2}}}} dx = \int {\frac{1}{{{4^2} - {x^2}}}} dx \cr
& {\text{The integrand has a expression in the form }}{a^2} - {u^2}{\text{ }}{\text{}} \cr
& {\text{Use formula 69}} \cr
& \left( {69} \right):\,\,\,\,\int {\frac{{du}}{{{a^2} - {u^2}}}} = \frac{1}{{2a}}\ln \left| {\frac{{u + a}}{{u - a}}} \right| + C\,\,\, \cr
& {\text{let }}u = x,\,\,\,a = 4 \cr
& \int {\frac{1}{{{4^2} - {x^2}}}} dx = \frac{1}{{2\left( 4 \right)}}\ln \left| {\frac{{x + 4}}{{x - 4}}} \right| + C\,\,\, \cr
& {\text{simplifying}} \cr
& \int {\frac{1}{{16 - {x^2}}}} dx = \frac{1}{8}\ln \left| {\frac{{x + 4}}{{x - 4}}} \right| + C\,\,\, \cr} $$
