Answer
$$\frac{1}{{2\sqrt 2 }}{\coth ^{ - 1}}\frac{x}{{2\sqrt 2 }} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{8 - {x^2}}}} , \cr
& {\text{rewriting the denominator}} \cr
& \int {\frac{{dx}}{{{{\left( {\sqrt 8 } \right)}^2} - {x^2}}}} \cr
& {\text{find the antiderivative using the theorem 6}}{\text{.12}}{\text{, with }}x{\text{ > 2}}\sqrt 2 \cr
& \int {\frac{{dx}}{{{{\left( {\sqrt 8 } \right)}^2} - {x^2}}} = \frac{1}{{\sqrt 8 }}{{\coth }^{ - 1}}\frac{x}{{\sqrt 8 }} + C} \cr
& \sqrt 8 = 2\sqrt 2 ,then \cr
& = \frac{1}{{2\sqrt 2 }}{\coth ^{ - 1}}\frac{x}{{2\sqrt 2 }} + C \cr} $$
