Answer
$$(\ln 4) \mathbf{i}+(\ln 4) \mathbf{j}+(\ln 2) \mathbf{k}$$
Work Step by Step
Given $$ \int_{1}^{4}\left(\frac{1}{t} \mathbf{i}+\frac{1}{5-t} \mathbf{j}+\frac{1}{2 t} \mathbf{k}\right) d t $$
Then
\begin{align*}
\int_{1}^{4}\left(\frac{1}{t} \mathbf{i}+\frac{1}{5-t} \mathbf{j}+\frac{1}{2 t} \mathbf{k}\right) d t&=[\ln t]\bigg|_{1}^{4} \mathbf{i}+[-\ln (5-t)]\bigg|_{1}^{4} \mathbf{j}+\left[\frac{1}{2} \ln t\right]\bigg|_{1}^{4} \mathbf{k}\\
&=(\ln 4) \mathbf{i}+(\ln 4) \mathbf{j}+(\ln 2) \mathbf{k}
\end{align*}