Answer
$$ - \frac{1}{4}{\operatorname{csch} ^{ - 1}}\frac{{\left| x \right|}}{4} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{x\sqrt {16 + {x^2}} }}} \cr
& {\text{rewriting }} \cr
& \int {\frac{{dx}}{{x\sqrt {{{\left( 4 \right)}^2} + {x^2}} }}} \cr
& {\text{find the denominator using the theorem 6}}{\text{.12}} \cr
& \int {\frac{{dx}}{{x\sqrt {{a^2} + {x^2}} }}} = - \frac{1}{a}{\operatorname{csch} ^{ - 1}}\frac{{\left| x \right|}}{a} + C \cr
& ,so \cr
& = - \frac{1}{4}{\operatorname{csch} ^{ - 1}}\frac{{\left| x \right|}}{4} + C \cr} $$
