Determine the integral of the given function:
$\int t^{3}\left(t^{4}-6\right)^{4} d t=$
Answer
The integral of the given function is \(\boxed{C + \frac{t^{14}}{56} - \frac{3t^{10}}{5} + 9t^{6} - 108t^{2} - \frac{162}{t^{2}}}\).
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 \(\boxed{C + \frac{t^{14}}{56} - \frac{3t^{10}}{5} + 9t^{6} - 108t^{2} - \frac{162}{t^{2}}}\).
