Answer
$$\int\left[\ln t \ \mathbf{i}+\frac{1}{t} \mathbf{j}+\mathbf{k}\right] d t=(t \ln t-t) \mathbf{i}+\ln t \mathbf{j}+t \mathbf{k}+\mathbf{C}$$
Work Step by Step
Given $$\int\left[\ln t \ \mathbf{i}+\frac{1}{t} \mathbf{j}+\mathbf{k}\right] d t $$
to find $$\int \ln t \ dt $$
let $u=\ln t, \ \ du=\frac{1}{t} dt \ \ \ \ and \ \ \ \ \ \ \ \ \ \ \ dv=dt , v=t$
this implies $$\int \ln t \ dt =u v- \int v du\\
=t \ln t-\int dt=t \ln t-t$$
Integrating on a component-by-component basis produces
\begin{align}\int\left[\ln t \mathbf{i}+\frac{1}{t} \mathbf{j}+\mathbf{k}\right] d t=(t \ln t-t) \mathbf{i}+\ln t \mathbf{j}+t \mathbf{k}+\mathbf{C}
\end{align}
