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