Answer
$$u\arctan u - \ln \sqrt {1 + {u^2}} + C$$
Work Step by Step
$$\eqalign{
& \int {\arctan u} du \cr
& {\text{Use integration by parts}} \cr
& t = \arctan u,{\text{ }}dt = \frac{1}{{1 + {u^2}}}du \cr
& dv = du,{\text{ }}v = u \cr
& \int {tdv} = tv - \int {vdt} \cr
& {\text{Substituting}} \cr
& \int {\arctan u} du = \left( {\arctan u} \right)\left( u \right) - \int u \left( {\frac{1}{{1 + {u^2}}}} \right)du \cr
& \int {\arctan u} du = u\arctan u - \int {\frac{u}{{1 + {u^2}}}} du \cr
& {\text{Rewrite}} \cr
& \int {\arctan u} du = u\arctan u - \frac{1}{2}\int {\frac{{2u}}{{1 + {u^2}}}} du \cr
& {\text{Recall that }}\int {\frac{{dt}}{t} = \ln \left| t \right| + C} \cr
& \int {\arctan u} du = u\arctan u - \frac{1}{2}\ln \left| {1 + {u^2}} \right| + C \cr
& = u\arctan u - \frac{1}{2}\ln \left( {1 + {u^2}} \right) + C \cr
& {\text{Use logarithmic properties}} \cr
& = u\arctan u - \ln \sqrt {1 + {u^2}} + C \cr} $$
