x⁶ + 1 = x²[(x⁴ + 1) - 1] + 1
= x²(x⁴ + 1) - x² + 1
∫ (x⁶ + 1)/(x⁴ + 1) dx
= ∫ [x²(x⁴ + 1) + 1 - x²]/(x⁴ + 1) dx
= ∫ x² dx - ∫ (x² - 1)/(x⁴ + 1) dx
= x³/3 - ∫ (1 - 1/x²)/(x² + 1/x²) dx
= x³/3 - ∫ d(x + 1/x)/[(x + 1/x)² - 2]
= x³/3 - 1/(2√2) * ln| [(x + 1/x) - √2]/[(x + 1/x) + √2] | + C
= x³/3 - (√2/4)ln| (x² - √2x + 1)/(x² + √2x + 1) | + C
再问： 分子跟分母搞反了。。。
再答： (x^4+1-x^2)(x^2+1)=x^6+1 ∫ (x^4+1)/(x^6+1)dx =∫ (x^4+1-x^2+x^2)/(x^6+1)dx =∫ (x^4+1-x^2)/(x^6+1)dx+∫ x^2/(x^6+1)dx =∫ 1/(x^2+1)dx+1/3∫ 1/(x^6+1)d(x^3) =arctanx+1/3arctan(x^3)+C
