Answer
$$y = \frac{1}{2}\sqrt {4 - {x^2}} - \ln \left| {\frac{{2 + \sqrt {4 - {x^2}} }}{x}} \right| + C$$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dx}} = \frac{{\sqrt {4 - {x^2}} }}{{2x}} \cr
& {\text{Separate the variables}} \cr
& dy = \frac{{\sqrt {4 - {x^2}} }}{{2x}}dx \cr
& {\text{Integrate both sides}} \cr
& \int {dy} = \int {\frac{{\sqrt {4 - {x^2}} }}{{2x}}} dx \cr
& y = \frac{1}{2}\int {\frac{{\sqrt {4 - {x^2}} }}{x}} dx \cr
& {\text{Integrate by tables, use }} \cr
& \int {\frac{{\sqrt {{a^2} - {u^2}} }}{u}du = } \sqrt {{a^2} - {u^2}} - a\ln \left| {\frac{{a + \sqrt {{a^2} - {u^2}} }}{u}} \right| + C \cr
& y = \frac{1}{2}\left( {\sqrt {4 - {x^2}} - 2\ln \left| {\frac{{2 + \sqrt {4 - {x^2}} }}{x}} \right|} \right) + C \cr
& y = \frac{1}{2}\sqrt {4 - {x^2}} - \ln \left| {\frac{{2 + \sqrt {4 - {x^2}} }}{x}} \right| + C \cr} $$
