Answer
$$\frac{1}{2}{\sin ^{ - 1}}x - 3{\tan ^{ - 1}}x + C$$
Work Step by Step
$$\eqalign{
& \int {\left[ {\frac{1}{{2\sqrt {1 - {x^2}} }} - \frac{3}{{1 + {x^2}}}} \right]} dx \cr
& {\text{Split the integrand}} \cr
& = \int {\frac{1}{{2\sqrt {1 - {x^2}} }}} dx - \int {\frac{3}{{1 + {x^2}}}} dx \cr
& = \frac{1}{2}\int {\frac{1}{{\sqrt {1 - {x^2}} }}} dx - 3\int {\frac{1}{{1 + {x^2}}}} dx \cr
& {\text{Use the rule of integration for inverse trigonometric functions}} \cr
& = \frac{1}{2}{\sin ^{ - 1}}x - 3{\tan ^{ - 1}}x + C \cr
& \cr
& \cr
& {\text{Check by differentiation}} \cr
& \frac{d}{{dx}}\left[ {\frac{1}{2}{{\sin }^{ - 1}}x - 3{{\tan }^{ - 1}}x + C} \right] \cr
& = \frac{1}{2}\left( {\frac{1}{{\sqrt {1 - {x^2}} }}} \right) - 3\left( {\frac{1}{{1 + {x^2}}}} \right) + 0 \cr
& = \frac{1}{{2\sqrt {1 - {x^2}} }} - \frac{3}{{1 + {x^2}}} \cr} $$
