Answer
$$\mathbf{r}(t)= (-\frac{1}{2} e^{-t^{2}}+1) \mathbf{i}+(e^{-t} -2)\mathbf{j}+(t +1)\ \mathbf{k}$$
Work Step by Step
Given$$ \mathbf{r}^{\prime}(t)=t e^{-t^{2}} \mathbf{i}-e^{-t} \mathbf{j}+\mathbf{k}, \quad \mathbf{r}(0)=\frac{1}{2} \mathbf{i}-\mathbf{j}+\mathbf{k}$$
As we have $$\mathbf{r}(t)=\int \mathbf{ r^{\prime}}(t)dt$$
So, integrating on a component-by-component basis produces
\begin{array}{l}{\mathbf{r}(t)=\int \mathbf{ r^{\prime}}(t)dt=\int\left(t e^{-t^{2}} \mathbf{i}-e^{-t} \mathbf{j}+\mathbf{k}\right) d t\\
=-\frac{1}{2} e^{-t^{2}} \mathbf{i}+e^{-t} \mathbf{j}+t \ \mathbf{k}+\mathbf{C}} \\
\text{since} \ \mathbf{r}(0)=\frac{1}{2} \mathbf{i}-\mathbf{j}+\mathbf{k}, \quad \text{so we get}
\\ {\mathbf{r}(0)=-\frac{1}{2} e^{0} \mathbf{i}+e^{0} \mathbf{j}+0 \ \mathbf{k}+\mathbf{C}=\frac{1}{2} \mathbf{i}-\mathbf{j}+\mathbf{k}}
\\ {\mathbf{r}(0)=-\frac{1}{2} \mathbf{i}+ \mathbf{j}+0 \ \mathbf{k}+\mathbf{C}=\frac{1}{2} \mathbf{i}-\mathbf{j}+\mathbf{k}\\ \Rightarrow \mathbf{C}= \mathbf{i}-2\mathbf{j}+\mathbf{k}} \\
{\text {so we get }}\\
{\mathbf{r}(t)= -\frac{1}{2} e^{-t^{2}} \mathbf{i}+e^{-t} \mathbf{j}+t \ \mathbf{k}+\mathbf{i}-2\mathbf{j}+\mathbf{k}\\
=(-\frac{1}{2} e^{-t^{2}}+1) \mathbf{i}+(e^{-t} -2)\mathbf{j}+(t +1)\ \mathbf{k}}\end{array}
