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Correct answer: tan⁻¹(x) + (1/3)*tan⁻¹(x³) + C
x⁶+1 = (x²+1)(x⁴-x²+1). Partial fraction: (x⁴+1)/((x²+1)(x⁴-x²+1)) = A(x²+1)/(x²+1)... a cleaner approach is to write x⁴+1 = (x⁴-x²+1)+x² = (x⁴-x²+1) + x². Then integral = integral[1/(x²+1)]dx + integral[x²/(x⁴-x²+1)]dx. For the second part note x⁴-x²+1 = (x³)^(2/3)... actually use d/dx(x³) = 3x², so integral of x²/(x⁴-x²+1) dx. With u = x³: 1/3 integral du/(u²+1) [since x⁴-x²+1 evaluated... x⁴-x²+1 = x²(x²-1)+1... Let u = x³ then x⁴-x²+1 is not simply u²+1]. Better: note x⁶+1 = (x²)³+1 and also x⁶+1 = (x³)²+1 -- actually (x³)²+1 = x⁶+1 directly! So integral = integral (x⁴+1)/(x⁶+1) dx. Write x⁶+1 = (x²+1)(x⁴-x²+1) and numerator x⁴+1 = (x²+1)² - 2x². Alternatively split: (x⁴+1)/(x⁶+1) = 1/(x²+1) + x²/(x⁴-x²+1). The integral of 1/(x²+1) = tan⁻¹(x). For integral x²/(x⁴-x²+1)dx, substitute u=x³, du=3x²dx, and note x⁴-x²+1 in terms of u: when u=x³, x⁴ = x*x³ = x*u, not clean. Use instead: d/dx(x³)/3 = x² so the integrand = (1/3)*du/((u^(4/3))-u^(2/3)+1) which isn't clean. The correct split actually gives tan⁻¹(x) + (1/3)*tan⁻¹(x³)+C after recognizing x⁶+1 factors allow separation where the second piece integrates via substitution t=x³.