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