1. **State the problem:** We need to find the integral $$\int \frac{t^4}{t^3+1} \, dt.$$\n\n2. **Rewrite the integrand:** Notice that the degree of the numerator ($4$) is higher than the degree of the denominator ($3$), so we perform polynomial division.\n\n3. **Polynomial division:** Divide $t^4$ by $t^3+1$.\n$$t^4 \div (t^3+1) = t - \frac{t}{t^3+1}$$\nThis is because $t^4 = t \cdot (t^3+1) - t$.\n\n4. **Rewrite the integral:**\n$$\int \frac{t^4}{t^3+1} \, dt = \int \left(t - \frac{t}{t^3+1}\right) dt = \int t \, dt - \int \frac{t}{t^3+1} \, dt.$$\n\n5. **Integrate the first term:**\n$$\int t \, dt = \frac{t^2}{2} + C.$$\n\n6. **Focus on the second integral:** $$\int \frac{t}{t^3+1} \, dt.$$\nUse substitution: let $$u = t^3 + 1,$$ then $$du = 3t^2 dt.$$\nWe have $t dt$ in the numerator, but $du$ involves $t^2 dt$, so rewrite $t dt$ as $$t dt = \frac{du}{3t}$$ which is complicated. Instead, try partial fraction decomposition on $$\frac{t}{t^3+1}.$$\n\n7. **Factor denominator:**\n$$t^3 + 1 = (t+1)(t^2 - t + 1).$$\n\n8. **Set up partial fractions:**\n$$\frac{t}{(t+1)(t^2 - t + 1)} = \frac{A}{t+1} + \frac{Bt + C}{t^2 - t + 1}.$$\nMultiply both sides by denominator:\n$$t = A(t^2 - t + 1) + (Bt + C)(t+1).$$\n\n9. **Expand right side:**\n$$t = A t^2 - A t + A + B t^2 + B t + C t + C.$$\nGroup terms:\n$$t = (A + B) t^2 + (-A + B + C) t + (A + C).$$\n\n10. **Equate coefficients:**\nFor $t^2$: $$0 = A + B,$$\nFor $t$: $$1 = -A + B + C,$$\nFor constant: $$0 = A + C.$$\n\n11. **Solve system:**\nFrom $0 = A + B$, we get $$B = -A.$$\nFrom $0 = A + C$, we get $$C = -A.$$\nSubstitute into $1 = -A + B + C$: $$1 = -A + (-A) + (-A) = -3A \implies A = -\frac{1}{3}.$$\nThen $$B = -A = \frac{1}{3}, \quad C = -A = \frac{1}{3}.$$\n\n12. **Rewrite integral:**\n$$\int \frac{t}{t^3+1} dt = \int \left( \frac{-\frac{1}{3}}{t+1} + \frac{\frac{1}{3} t + \frac{1}{3}}{t^2 - t + 1} \right) dt = -\frac{1}{3} \int \frac{1}{t+1} dt + \frac{1}{3} \int \frac{t+1}{t^2 - t + 1} dt.$$\n\n13. **Integrate first term:**\n$$-\frac{1}{3} \int \frac{1}{t+1} dt = -\frac{1}{3} \ln|t+1| + C.$$\n\n14. **Integrate second term:**\nRewrite numerator: $$t+1 = \frac{1}{2} (2t - 1) + \frac{3}{2}.$$\nSo,\n$$\int \frac{t+1}{t^2 - t + 1} dt = \int \frac{\frac{1}{2} (2t - 1) + \frac{3}{2}}{t^2 - t + 1} dt = \frac{1}{2} \int \frac{2t - 1}{t^2 - t + 1} dt + \frac{3}{2} \int \frac{1}{t^2 - t + 1} dt.$$\n\n15. **First integral:** Let $$u = t^2 - t + 1,$$ then $$du = (2t - 1) dt.$$\nSo,\n$$\frac{1}{2} \int \frac{2t - 1}{t^2 - t + 1} dt = \frac{1}{2} \int \frac{du}{u} = \frac{1}{2} \ln|t^2 - t + 1| + C.$$\n\n16. **Second integral:** Complete the square in denominator:\n$$t^2 - t + 1 = \left(t - \frac{1}{2}\right)^2 + \frac{3}{4}.$$\nSo,\n$$\int \frac{1}{\left(t - \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} dt = \frac{2}{\sqrt{3}} \arctan \left( \frac{2t - 1}{\sqrt{3}} \right) + C.$$\n\n17. **Combine second integral:**\n$$\frac{3}{2} \times \frac{2}{\sqrt{3}} \arctan \left( \frac{2t - 1}{\sqrt{3}} \right) = \frac{3}{\sqrt{3}} \arctan \left( \frac{2t - 1}{\sqrt{3}} \right) = \sqrt{3} \arctan \left( \frac{2t - 1}{\sqrt{3}} \right).$$\n\n18. **Final integral result:**\n$$\int \frac{t}{t^3+1} dt = -\frac{1}{3} \ln|t+1| + \frac{1}{6} \ln|t^2 - t + 1| + \frac{\sqrt{3}}{3} \arctan \left( \frac{2t - 1}{\sqrt{3}} \right) + C.$$\n\n19. **Recall step 4:**\n$$\int \frac{t^4}{t^3+1} dt = \int t dt - \int \frac{t}{t^3+1} dt = \frac{t^2}{2} - \left(-\frac{1}{3} \ln|t+1| + \frac{1}{6} \ln|t^2 - t + 1| + \frac{\sqrt{3}}{3} \arctan \left( \frac{2t - 1}{\sqrt{3}} \right) \right) + C.$$\n\n20. **Simplify final answer:**\n$$\boxed{\frac{t^2}{2} + \frac{1}{3} \ln|t+1| - \frac{1}{6} \ln|t^2 - t + 1| - \frac{\sqrt{3}}{3} \arctan \left( \frac{2t - 1}{\sqrt{3}} \right) + C}.$$