Answer
$$\frac{{2{{\left( {{{\tan }^{ - 1}}x} \right)}^{3/2}}}}{3} + C $$
Work Step by Step
$$\eqalign{
& \int {\frac{{\sqrt {{{\tan }^{ - 1}}x} dx}}{{1 + {x^2}}}} \cr
& {\text{use the substitution method}}{\text{.}} \cr
& u = {\tan ^{ - 1}}x,{\text{ so that }}du = \frac{1}{{1 + {x^2}}}dx \cr
& {\text{then}} \cr
& \int {\frac{{\sqrt {{{\tan }^{ - 1}}x} dx}}{{1 + {x^2}}}} = \int {\sqrt u } du \cr
& = \int {{u^{1/2}}} du \cr
& {\text{Integrating by the power rule}} \cr
& = \frac{{{u^{3/2}}}}{{3/2}} + C \cr
& = \frac{{2{u^{3/2}}}}{3} + C \cr
& {\text{write in terms of }}x;{\text{ replace }}{\tan ^{ - 1}}x{\text{ for }}u \cr
& = \frac{{2{{\left( {{{\tan }^{ - 1}}x} \right)}^{3/2}}}}{3} + C \cr} $$
