Answer
$$x + {\tan ^{ - 1}}x + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{2 + {x^2}}}{{1 + {x^2}}}dx} \cr
& = \int {\frac{{1 + {x^2} + 1}}{{1 + {x^2}}}dx} \cr
& {\text{by the long division}} \cr
& = \int {\left( {1 + \frac{1}{{1 + {x^2}}}} \right)dx} \cr
& {\text{split the integrand}} \cr
& = \int {dx} + \int {\frac{1}{{1 + {x^2}}}} dx \cr
& {\text{integrate}} \cr
& = x + {\tan ^{ - 1}}x + C \cr
& {\text{check by differentiation}} \cr
& {\text{ = }}\frac{d}{{dx}}\left( {x + {{\tan }^{ - 1}}x + C} \right) \cr
& {\text{ = }}\frac{d}{{dx}}\left( x \right) + \frac{d}{{dx}}\left( {{{\tan }^{ - 1}}x} \right) + \frac{d}{{dx}}\left( C \right) \cr
& {\text{ = }}1 + \frac{1}{{1 + {x^2}}} + 0 \cr
& {\text{simplify}} \cr
& {\text{ = }}\frac{{1 + {x^2} + 1}}{{1 + {x^2}}} \cr
& add \cr
& = \frac{{2 + {x^2}}}{{1 + {x^2}}} \cr} $$
