To find the indefinite integral of the vector function $(6t\mathbf{i} + \mathbf{j} + \mathbf{k}) \, dt$, we integrate each component of the vector with respect to $t$.

The vector function can be broken into its components:

$6t\mathbf{i} + \mathbf{j} + \mathbf{k}$

1. Integral of $6t\mathbf{i}$: $\int 6t \, dt = 6 \cdot \frac{t^2}{2} = 3t^2$ So, the integral of $6t\mathbf{i}$ is $3t^2\mathbf{i}$.

2. Integral of $\mathbf{j}$: $\int 1 \, dt = t$ So, the integral of $\mathbf{j}$ is $t\mathbf{j}$.

3. Integral of $\mathbf{k}$: $\int 1 \, dt = t$ So, the integral of $\mathbf{k}$ is $t\mathbf{k}$.

Final result:

$\int (6t\mathbf{i} + \mathbf{j} + \mathbf{k}) \, dt = 3t^2\mathbf{i} + t\mathbf{j} + t\mathbf{k} + \mathbf{C}$

where $\mathbf{C}$ is the constant of integration, a vector $(C_1\mathbf{i} + C_2\mathbf{j} + C_3\mathbf{k})$.

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Here are 5 related questions you can explore:

1. How do you find the indefinite integral of a vector function?
2. What is the difference between definite and indefinite integrals in vector calculus?
3. How do constants of integration work in vector calculus?
4. How would you compute a line integral for a vector field?
5. What is the geometric interpretation of integrating a vector function?

Tip: When integrating vector functions, treat each component separately, integrating them as scalar functions of the variable.