Answer
$$\int\left[(2 t-1) \mathbf{i}+4 t^{3} \mathbf{j}+3 \sqrt{t} \ \mathbf{k}\right] d t=\left(t^{2}-t\right) \mathbf{i}+t^{4} \mathbf{j}+2 t^{3 / 2} \mathbf{k}+\mathbf{C}$$
Work Step by Step
Given $$\int\left[(2 t-1) \mathbf{i}+4 t^{3} \mathbf{j}+3 \sqrt{t} \ \mathbf{k}\right] d t $$
Integrating on a component-by-component basis produces
\begin{align}
I&=\int\left[(2 t-1) \mathbf{i}+4 t^{3} \mathbf{j}+3 \sqrt{t} \ \mathbf{k}\right] d t\\
&=\left(\frac{2}{2}t^{2}-t\right) \mathbf{i}+\frac{4}{4}t^{4} \mathbf{j}+\frac{3}{3/2} t^{\frac{1 }{ 2}+1} \mathbf{k}+\mathbf{C}\\
&=\left(t^{2}-t\right) \mathbf{i}+t^{4} \mathbf{j}+2 t^{3 / 2} \mathbf{k}+\mathbf{C}
\end{align}
