Answer
$$\eqalign{
& {\bf{r}}\left( t \right) = \left\langle {{e^t}, - \cos t,\tan t} \right\rangle + \left\langle {1,3,2} \right\rangle \cr
& {\bf{r}}\left( t \right) = \left\langle {{e^t} + 1,3 - \cos t,\tan t + 2} \right\rangle \cr} $$
Work Step by Step
$$\eqalign{
& {\bf{r}}'\left( t \right) = \left\langle {{e^t},\sin t,{{\sec }^2}t} \right\rangle ;\,\,\,\,\,\,{\bf{r}}\left( 0 \right) = \left\langle {2,2,2} \right\rangle \cr
& \int {{\bf{r}}'\left( t \right)dt} = \int {\left\langle {{e^t},\sin t,{{\sec }^2}t} \right\rangle dt} \cr
& {\bf{r}}\left( t \right) = \left\langle {{e^t}, - \cos t,\tan t} \right\rangle + {\bf{C}}\,\,\,\left( 1 \right) \cr
& \cr
& {\text{Use the initial condition}} \cr
& \left\langle {2,2,2} \right\rangle = \left\langle {{e^0}, - \cos 0,\tan 0} \right\rangle + {\bf{C}} \cr
& \left\langle {2,2,2} \right\rangle = \left\langle {1, - 1,0} \right\rangle + {\bf{C}} \cr
& {\bf{C}} = \left\langle {2,2,2} \right\rangle - \left\langle {1, - 1,0} \right\rangle \cr
& {\bf{C}} = \left\langle {2 - 1,2 + 1,2 - 0} \right\rangle \cr
& {\bf{C}} = \left\langle {1,3,2} \right\rangle \cr
& \cr
& {\text{Substitute the vector constant into }}\left( 1 \right) \cr
& {\bf{r}}\left( t \right) = \left\langle {{e^t}, - \cos t,\tan t} \right\rangle + \left\langle {1,3,2} \right\rangle \cr
& {\bf{r}}\left( t \right) = \left\langle {{e^t} + 1,3 - \cos t,\tan t + 2} \right\rangle \cr} $$
