Question

∫x^3dx/(x^4-x^2+2)

Answer 1

∫ x³/(x⁴-x²+2) dx
=(1/2)∫ x²/(x⁴-x²+2) d(x²)
令x²=u
=(1/2)∫ u/(u²-u+2) du
=(1/2)∫ (u-1/2+1/2)/(u²-u+2) du
=(1/4)∫ (2u-1)/(u²-u+2) du + (1/4)∫ 1/(u²-u+2) du
=(1/4)∫ 1/(u²-u+2) d(u²-u) + (1/4)∫ 1/[(u-1/2)²+7/4] du
=(1/4)ln(u²-u+2) + (√7/14)arctan[2(u-1/2)/√7] + C
=(1/4)ln(x⁴-x²+2) + (√7/14)arctan[(2x²-1)/√7] + C

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Answer 2

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