Answer
$${\bf{r}}\left( t \right) = \left\langle {t + 3,{t^2} + 2,{t^3} - 6} \right\rangle $$
Work Step by Step
$$\eqalign{
& {\bf{r}}'\left( t \right) = \left\langle {1,2t,3{t^2}} \right\rangle ;\,\,\,\,\,\,{\bf{r}}\left( 1 \right) = \left\langle {4,3, - 5} \right\rangle \cr
& \int {{\bf{r}}'\left( t \right)dt} = \int {\left\langle {1,2t,3{t^2}} \right\rangle dt} \cr
& {\bf{r}}\left( t \right) = \left\langle {t,{t^2},{t^3}} \right\rangle + {\bf{C}}\,\,\,\left( 1 \right) \cr
& \cr
& {\text{Use the initial condition}} \cr
& \left\langle {4,3, - 5} \right\rangle = \left\langle {{{1,1}^2}{{,1}^3}} \right\rangle + {\bf{C}} \cr
& \left\langle {4,3, - 5} \right\rangle = \left\langle {1,1,1} \right\rangle + {\bf{C}} \cr
& {\bf{C}} = \left\langle {4,3, - 5} \right\rangle - \left\langle {1,1,1} \right\rangle \cr
& {\bf{C}} = \left\langle {3,2, - 6} \right\rangle \cr
& \cr
& {\text{Substitute the vector constant into }}\left( 1 \right) \cr
& {\bf{r}}\left( t \right) = \left\langle {t,{t^2},{t^3}} \right\rangle + \left\langle {3,2, - 6} \right\rangle \cr
& {\bf{r}}\left( t \right) = \left\langle {t + 3,{t^2} + 2,{t^3} - 6} \right\rangle \cr} $$
