Answer
$$3(\ln 3-1) \mathbf{i}+(3-e) \mathbf{j}+(\ln 3(\ln (\ln 3)-1)+1) \mathbf{k} $$
Work Step by Step
We evaluate the integral of the vector function as follows:
\begin{align*}
\int_{1}^{\ln 3}\left(t e^{t} \mathbf{i}+e^{t} \mathbf{j}+\ln t \mathbf{k}\right) d t&= \left( te^t\bigg|_{1}^{\ln 3} - \int_{1}^{\ln 3} e^tdt\right)\mathbf{i}-\left[e^{t}\right]_{1}^{\ln 3} \mathbf{j}+[t \ln t-t]_{1}^{\ln 3} \mathbf{k} \\
&=\left[t e^{t}-e^{t}\right]\bigg|_{1}^{\ln 3} \mathbf{i}-\left[e^{t}\right]\bigg|_{1}^{\ln 3} \mathbf{j}+[t \ln t-t]\bigg|_{1}^{\ln 3} \mathbf{k} \\
&=3(\ln 3-1) \mathbf{i}+(3-e) \mathbf{j}+(\ln 3(\ln (\ln 3)-1)+1) \mathbf{k}
\end{align*}
