Determine the integral of the given function:
$\int t^3{(t^4-6)}^{4}dt=$

Step 1 ：Let's start by using the substitution method. We let $u=t^4-6$.

Step 2 ：Then, we find the derivative of $u$ with respect to $t$, which gives us $du=4t^{3}dt$.

Step 3 ：We can now rewrite the integral in terms of $u$ and solve it.

Step 4 ：After solving, we get $\frac{t^{14}}{56}-\frac{3t^{10}}{5}+9t^6-108t^2-\frac{162}{t^2}$.

Step 5 ：However, the integral of a function is not unique; adding any constant to it will still result in the same derivative. Therefore, we add the constant of integration, usually denoted as 'C', to the result.

Step 6 ：Finally, we substitute $u$ back with $t^4-6$ to get the final answer.

Step 7 ：The integral of the given function is $\boxed{{\displaystyle C+\frac{t^{14}}{56}-\frac{3t^{10}}{5}+9t^6-108t^2-\frac{162}{t^2}}}$.