Answer
$${\tan ^{ - 1}}x - 2{\sin ^{ - 1}}x + C$$
Work Step by Step
$$\eqalign{
& \int {\left[ {\frac{1}{{1 + {x^2}}} + \frac{2}{{\sqrt {1 - {x^2}} }}} \right]} dx \cr
& {\text{sum rule}} \cr
& = \int {\frac{1}{{1 + {x^2}}}} dx - \int {\frac{2}{{\sqrt {1 - {x^2}} }}} dx \cr
& {\text{multiple constant rule}} \cr
& = \int {\frac{1}{{1 + {x^2}}}} dx - 2\int {\frac{1}{{\sqrt {1 - {x^2}} }}} dx \cr
& {\text{integrate by using the rules}} \cr
& \int {\frac{1}{{1 + {x^2}}}dx} = {\tan ^{ - 1}}x + C{\text{ and }}\int {\frac{1}{{\sqrt {1 - {x^2}} }}} dx = {\sin ^{ - 1}}x + C \cr
& = {\tan ^{ - 1}}x - 2{\sin ^{ - 1}}x + C \cr} $$
