Answer
$$\frac{e-1}{2} \mathbf{i}+\frac{e-1}{e} \mathbf{i}+\mathbf{k}$$
Work Step by Step
We evaluate the integral of the vector function as follows:
\begin{align*}
\int_{0}^{1}\left(t e^{t^{2}} \mathbf{i}+e^{-t} \mathbf{j}+\mathbf{k}\right) d t&=\left(\frac{1}{2} e^{t^{2}}\right)\bigg|_{0}^{1} \mathbf{i}-\left(e^{-t}\right)\bigg|_{0}^{1} \mathbf{j}+t\bigg|_{0}^{1} \mathbf{k}\\
&=\frac{e-1}{2} \mathbf{i}+\frac{e-1}{e} \mathbf{i}+\mathbf{k}
\end{align*}