Answer
$\frac{{{{\left( {{{\tan }^{ - 1}}x} \right)}^6}}}{6} + C$
Work Step by Step
$$\eqalign{
& \int {\frac{{{{\left( {{{\tan }^{ - 1}}x} \right)}^5}}}{{1 + {x^2}}}} dx \cr
& {\text{set }}u = {\tan ^{ - 1}}x{\text{ then }}du = \frac{1}{{1 + {x^2}}} \cr
& {\text{use the change of variable}} \cr
& \int {{{\left( {{{\tan }^{ - 1}}x} \right)}^5}\left( {\frac{1}{{1 + {x^2}}}} \right)} dx = \int {{u^5}} du \cr
& {\text{integrate}} \cr
& = \frac{{{u^6}}}{6} + C \cr
& {\text{replace }}u = {\tan ^{ - 1}}x \cr
& = \frac{{{{\left( {{{\tan }^{ - 1}}x} \right)}^6}}}{6} + C \cr} $$
