1. **Problem statement:** Find the indefinite integral $$\int \frac{1}{t^4} \, dt$$. 2. **Rewrite the integrand:** Recall that $$\frac{1}{t^4} = t^{-4}$$. 3. **Use the power rule for integration:** For any real number $n \neq -1$, $$\int t^n \, dt = \frac{t^{n+1}}{n+1} + C$$ 4. **Apply the rule:** Here, $n = -4$, so $$\int t^{-4} \, dt = \frac{t^{-4+1}}{-4+1} + C = \frac{t^{-3}}{-3} + C$$ 5. **Simplify the expression:** $$\frac{t^{-3}}{-3} = -\frac{1}{3t^3}$$ 6. **Final answer:** $$\int \frac{1}{t^4} \, dt = -\frac{1}{3t^3} + C$$ This matches the option "-1/3r3 + C" if we interpret $r$ as $t$.