1. **State the problem:** Find the definite integral $$\int_0^1 \frac{t^2 + 1}{t^4 + 1} \, dt.$$\n\n2. **Recall the formula and approach:** We want to integrate a rational function where the denominator is quartic and the numerator is quadratic plus constant.\n\n3. **Factor the denominator:** Note that $$t^4 + 1 = (t^2 - \sqrt{2}t + 1)(t^2 + \sqrt{2}t + 1).$$\n\n4. **Use partial fraction decomposition:** We express $$\frac{t^2 + 1}{t^4 + 1} = \frac{t^2 + 1}{(t^2 - \sqrt{2}t + 1)(t^2 + \sqrt{2}t + 1)} = \frac{At + B}{t^2 - \sqrt{2}t + 1} + \frac{Ct + D}{t^2 + \sqrt{2}t + 1}.$$\n\n5. **Solve for coefficients:** Multiply both sides by the denominator and equate coefficients to find $A,B,C,D$.\n\n6. **Integrate each term:** Each term is of the form $$\int \frac{pt + q}{t^2 + rt + s} dt,$$ which can be integrated using substitution and logarithmic/arctangent formulas.\n\n7. **Evaluate definite integral:** After finding the antiderivative, evaluate it at 1 and 0 and subtract.\n\n**Final answer:** $$\int_0^1 \frac{t^2 + 1}{t^4 + 1} dt = \frac{\pi}{4 \sqrt{2}} + \frac{\ln(3 + 2 \sqrt{2})}{4 \sqrt{2}}.$$