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用速解法。令u = x²，v = e^(3x)，对u求导，对v积分，然后交叉相乘，第一项正号，第二项负号，第三项正号...u' = 2x，v1 = (1/3)e^(3x)u'' = 2，v2 = (1/9)e^(3x)u''' = 0，v3 = (1/27)e^(3x)∫ x²e^(3x) dx= x² * (1/3)e^(3x) - 2x * (1/9)e^(3x) + 2 * (1/27)e^(3x) + C= (1/3)x²e^(3x) - (2/9)xe^(3x) + (2/27)e^(3x) + C 详细的则是分部积分法。∫ x²e^(3x) dx= (1/3)∫ x² d[e^(3x)]= (1/3)x²e^(3x) - (1/3)∫ 2xe^(3x) dx，第一次分部积分法= (1/3)x²e^(3x) - (2/3)(1/3)∫ x d[e^(3x)]= (1/3)x²e^(3x) - (2/9)xe^(3x) + (2/9)∫ e^(3x) dx，第二次分部积分法= (1/3)x²e^(3x) - (2/9)xe^(3x) + (2/9)(1/3)e^(3x) + C= (1/3)x²e^(3x) - (2/9)xe^(3x) + (2/27)e^(3x) + C
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