第1个回答 2021-07-29
令a=1即可，详情如图所示
第2个回答 2014-12-08
∫ x²arctanx dx
= ∫ arctanx d(x³/3)
= (1/3)x³arctanx - (1/3)∫ x³ d(arctanx)
= (1/3)x³arctanx - (1/3)∫ x³/(1 + x²) dx
= (1/3)x³arctanx - (1/3)∫ x[(1 + x²) - 1]/(1 + x²) dx
= (1/3)x³arctanx - (1/3)∫ [x - x/(1 + x²)] dx
= (1/3)x³arctanx - (1/3)(x²/2) + (1/3)(1/2)ln(1 + x²) + C
= (1/3)x³arctanx - x²/6 + (1/6)ln(1 + x²) + C追问
= ∫ arctanx d(x³/3)
= (1/3)x³arctanx - (1/3)∫ x³ d(arctanx)
= (1/3)x³arctanx - (1/3)∫ x³/(1 + x²) dx
= (1/3)x³arctanx - (1/3)∫ x[(1 + x²) - 1]/(1 + x²) dx
= (1/3)x³arctanx - (1/3)∫ [x - x/(1 + x²)] dx
= (1/3)x³arctanx - (1/3)(x²/2) + (1/3)(1/2)ln(1 + x²) + C
= (1/3)x³arctanx - x²/6 + (1/6)ln(1 + x²) + C追问
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