Answer
$$\int\left(4 t^{3} \mathbf{i}+6 \mathfrak{t} \mathbf{j}-4 \sqrt{t }\ \ \mathbf{k}\right) d t=t^{4} \mathbf{i}+3 t^{2} \mathbf{j}-\frac{8}{3} t^{\frac{3}{2}} \mathbf{k}+C$$
Work Step by Step
Given $$\int\left(4 t^{3} \mathbf{i}+6 \mathfrak{t} \mathbf{j}-4 \sqrt{t \mathbf{k}}\right) d t $$
Integrating on a component-by-component basis produces
\begin{align}\int\left(4 t^{3} \mathbf{i}+6 \mathfrak{t} \mathbf{j}-4 \sqrt{t }\ \ \mathbf{k}\right) d t&=\frac{4}{4}t^{4} \mathbf{i}+\frac{6}{2} t^{2} \mathbf{j}-\frac{4}{3/2} t^{\frac{3}{2}} \mathbf{k}+C\\
&= t^{4} \mathbf{i}+3 t^{2} \mathbf{j}-\frac{8}{3} t^{\frac{3}{2}} \mathbf{k}+C
\end{align}
