Answer
$$\sqrt {4 - {x^2}} - 2\ln \left| {\frac{{2 + \sqrt {4 - {x^2}} }}{x}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{\sqrt {4 - {x^2}} }}{x}} dx \cr
& {\text{Use the Endpaper Integral Table to evaluate the integral}} \cr
& {\text{Rewrite the integrand}} \cr
& \int {\frac{{\sqrt {4 - {x^2}} }}{x}} dx = \int {\frac{{\sqrt {{2^2} - {x^2}} }}{x}} dx \cr
& {\text{The integrand has a expression in the form }}\sqrt {{a^2} - {u^2}} {} \cr
& {\text{Use formula 79}} \cr
& \left( {80} \right):\,\,\,\,\int {\frac{{\sqrt {{a^2} - {u^2}} }}{u}du} = \sqrt {{a^2} - {u^2}} - a\ln \left| {\frac{{a + \sqrt {{a^2} - {u^2}} }}{u}} \right| + C \cr
& {\text{let }}u = x,\,\,\,a = 2 \cr
& \int {\frac{{\sqrt {{2^2} - {x^2}} }}{x}} dx = \sqrt {{2^2} - {x^2}} - 2\ln \left| {\frac{{2 + \sqrt {{2^2} - {x^2}} }}{x}} \right| + C \cr
& {\text{simplifying}} \cr
& \int {\frac{{\sqrt {4 - {x^2}} }}{x}} dx = \sqrt {4 - {x^2}} - 2\ln \left| {\frac{{2 + \sqrt {4 - {x^2}} }}{x}} \right| + C \cr} $$
