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