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