Answer
$$ y = {\tan ^{ - 1}}x - 2{\sin ^{ - 1}}x + 2$$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dx}} = \frac{1}{{1 + {x^2}}} - \frac{2}{{\sqrt {1 - {x^2}} }},\,\,\,\,\,\,\,y\left( 0 \right) = 2 \cr
& {\text{separate the variables}} \cr
& dy = \left( {\frac{1}{{1 + {x^2}}} - \frac{2}{{\sqrt {1 - {x^2}} }}} \right)dx \cr
& {\text{integrate both sides}} \cr
& y = \int {\left( {\frac{1}{{1 + {x^2}}} - \frac{2}{{\sqrt {1 - {x^2}} }}} \right)} dx \cr
& y = {\tan ^{ - 1}}x - 2{\sin ^{ - 1}}x + C\,\,\,\left( {\bf{1}} \right) \cr
& \cr
& {\text{use initial condition }}y\left( 0 \right) = 2 \cr
& 2 = {\tan ^{ - 1}}0 - 2{\sin ^{ - 1}}0 + C \cr
& C = 2 \cr
& \cr
& {\text{then}}{\text{, substitute }}C = 2{\text{ in }}\left( {\bf{1}} \right) \cr
& y = {\tan ^{ - 1}}x - 2{\sin ^{ - 1}}x + 2 \cr} $$
