Evaluate the integral. (Remember the constant of integration.) \[ \int t e^{-8 t} d t \]

Step 1 ：Given the integral \(\int t e^{-8 t} dt\), we will use the method of integration by parts to solve it.

Step 2 ：The formula for integration by parts is \(\int u dv = uv - \int v du\).

Step 3 ：We let \(u = t\) and \(dv = e^{-8t} dt\).

Step 4 ：We then find \(du\) and \(v\). In this case, \(du = 1\) and \(v = -\frac{1}{8}e^{-8t}\).

Step 5 ：Substituting these into the integration by parts formula, we get \(-\frac{t}{8}e^{-8t} - \int -\frac{1}{8}e^{-8t} dt\).

Step 6 ：Evaluating the remaining integral, we get \(-\frac{1}{64}e^{-8t}\).

Step 7 ：So, the integral of the function is \(-\frac{t}{8}e^{-8t} - \frac{1}{64}e^{-8t}\).

Step 8 ：However, we must remember to add the constant of integration, \(C\), to our final answer.

Step 9 ：Thus, the integral of \(t e^{-8 t}\) with respect to \(t\) is \(\boxed{-\frac{t}{8}e^{-8t} - \frac{1}{64}e^{-8t} + C}\).