求x*arctan根号xdx的不定积分 5
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简单计算一下即可，答案如图所示
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∫ x * arctan√x dx
= ∫ arctan√x d(x²/2)
= (x²/2)arctan√x - (1/2)∫ x² d(arctan√x)，分部积分法
= (x²/2)arctan√x - (1/2)∫ x² * 1/(1 + x) * 1/2√x dx
= (x²/2)arctan√x - (1/4)∫ x^(3/2)/(1 + x) dx，令u² = x，2udu = dx
= (x²/2)arctan√x - (1/2)∫ u^4/(1 + u²) du
= (x²/2)arctan√x - (1/2)∫ [u² - 1 + 1/(1 + u²)] du
= (x²/2)arctan√x - (1/2)(u³/3) + (1/2)u - (1/2)arctan(u) + C
= (x²/2)arctan√x - (1/6)x^(3/2) + (1/2)√x - (1/2)arctan√x + C
= (1/2)(x² - 1)arctan√x - (1/6)(x - 3)√x + C
= ∫ arctan√x d(x²/2)
= (x²/2)arctan√x - (1/2)∫ x² d(arctan√x)，分部积分法
= (x²/2)arctan√x - (1/2)∫ x² * 1/(1 + x) * 1/2√x dx
= (x²/2)arctan√x - (1/4)∫ x^(3/2)/(1 + x) dx，令u² = x，2udu = dx
= (x²/2)arctan√x - (1/2)∫ u^4/(1 + u²) du
= (x²/2)arctan√x - (1/2)∫ [u² - 1 + 1/(1 + u²)] du
= (x²/2)arctan√x - (1/2)(u³/3) + (1/2)u - (1/2)arctan(u) + C
= (x²/2)arctan√x - (1/6)x^(3/2) + (1/2)√x - (1/2)arctan√x + C
= (1/2)(x² - 1)arctan√x - (1/6)(x - 3)√x + C
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