Answer
$${\bf{r}}\left( t \right) = \left\langle {4,2t - 1,{t^2} - 6} \right\rangle $$
Work Step by Step
$$\eqalign{
& {\bf{r}}'\left( t \right) = \left\langle {0,2,2t} \right\rangle ;\,\,\,\,\,\,{\bf{r}}\left( 1 \right) = \left\langle {4,3, - 5} \right\rangle \cr
& \int {{\bf{r}}'\left( t \right)dt} = \int {\left\langle {0,2,2t} \right\rangle dt} \cr
& {\bf{r}}\left( t \right) = \left\langle {k,2t,{t^2}} \right\rangle + {\bf{C}}\,\,\,\left( 1 \right) \cr
& \cr
& {\text{Use the initial condition}} \cr
& \left\langle {4,3, - 5} \right\rangle = \left\langle {k,2\left( 1 \right),{{\left( 1 \right)}^2}} \right\rangle + {\bf{C}} \cr
& \left\langle {4,3, - 5} \right\rangle = \left\langle {k,2,1} \right\rangle + {\bf{C}} \cr
& {\bf{C}} = \left\langle {4,3, - 5} \right\rangle - \left\langle {k,2,1} \right\rangle \cr
& {\bf{C}} = \left\langle {4 - k,3 - 2, - 5 - 1} \right\rangle \cr
& {\bf{C}} = \left\langle {4 - k,1, - 6} \right\rangle \cr
& \cr
& {\text{Substitute the vector constant into }}\left( 1 \right) \cr
& {\bf{r}}\left( t \right) = \left\langle {k,2t,{t^2}} \right\rangle + \left\langle {4 - k,1, - 6} \right\rangle \cr
& {\bf{r}}\left( t \right) = \left\langle {k + 4 - k,2t + 1,{t^2} - 6} \right\rangle \cr
& {\bf{r}}\left( t \right) = \left\langle {4,2t - 1,{t^2} - 6} \right\rangle \cr} $$