Answer
$${\tan ^{ - 1}}\left( {\ln t} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{1}{{t\left[ {1 + {{\left( {\ln t} \right)}^2}} \right]}}} dt \cr
& {\text{Let }}u = \ln t.{\text{ Then }}du = \frac{1}{t}dt,{\text{ }}dt = tdu{\text{ substituting you have}} \cr
& \int {\frac{1}{{t\left[ {1 + {{\left( {\ln t} \right)}^2}} \right]}}} dt = \int {\frac{1}{{t\left[ {1 + {u^2}} \right]}}} \left( {tdu} \right) \cr
& = \int {\frac{1}{{1 + {u^2}}}} du \cr
& {\text{From the table of integrals in the back of the book}} \cr
& \int {\frac{1}{{{a^2} + {u^2}}}} du = \frac{1}{a}{\tan ^{ - 1}}\left( {\frac{u}{a}} \right) + C \cr
& \int {\frac{1}{{1 + {u^2}}}} du = {\tan ^{ - 1}}u + C \cr
& {\text{Write in terms of }}t,{\text{ }}u = \ln t \cr
& = {\tan ^{ - 1}}\left( {\ln t} \right) + C \cr} $$
