Answer
$${\bf{r}}\left( t \right) = \ln \left| {\sec t + \tan t} \right|{\bf{i}} - \ln \left| {\cos t} \right|{\bf{j}} + \left( {\frac{1}{3}{t^3} + 3} \right){\bf{k}}$$
Work Step by Step
$$\eqalign{
& {\bf{r}}'\left( t \right) = \sec t{\bf{i}} + \tan t{\bf{j}} + {t^2}{\bf{k}} \cr
& {\text{Integrating}} \cr
& {\bf{r}}\left( t \right) = \int {{\bf{r}}'\left( t \right)} dt \cr
& = \left[ {\int {\sec tdt} } \right]{\bf{i}} + \left[ {\int {\tan tdt} } \right]{\bf{j}} + \left[ {\int {{t^2}dt} } \right]{\bf{k}} \cr
& {\bf{r}}\left( t \right) = \ln \left| {\sec t + \tan t} \right|{\bf{i}} - \ln \left| {\cos t} \right|{\bf{j}} + \frac{1}{3}{t^3}{\bf{k}} + {\bf{C}} \cr
& {\text{Using the fact that }}{\bf{r}}\left( 0 \right) = 3{\bf{k}} \cr
& {\bf{r}}\left( 0 \right) = \ln \left| {\sec 0 + \tan 0} \right|{\bf{i}} - \ln \left| {\cos 0} \right|{\bf{j}} + \frac{1}{3}{\left( 0 \right)^3}{\bf{k}} + {\bf{C}} \cr
& {\bf{r}}\left( 0 \right) = \left( 0 \right){\bf{i}} - \left( 0 \right){\bf{j}} + 0{\bf{k}} + {\bf{C}} \cr
& 3{\bf{k}} = {\bf{C}} \cr
& {\text{Therefore,}} \cr
& {\bf{r}}\left( t \right) = \ln \left| {\sec t + \tan t} \right|{\bf{i}} - \ln \left| {\cos t} \right|{\bf{j}} + \frac{1}{3}{t^3}{\bf{k}} + 3{\bf{k}} \cr
& {\bf{r}}\left( t \right) = \ln \left| {\sec t + \tan t} \right|{\bf{i}} - \ln \left| {\cos t} \right|{\bf{j}} + \left( {\frac{1}{3}{t^3} + 3} \right){\bf{k}} \cr} $$
