標題:
發問:
integral of sqrt(x)*log(x) in到去(2/3)x^3/2 ln(x)-(4/9)(x^3/2)就唔知點做到去 (2/9)x^3/2 (3ln(x)-2) http://img40.imageshack.us/img40/6935/msp303591a493467i15adii.png
最佳解答:
= (2/3)ln(x)x^(3/2) - (4/9)x^(3/2) + C 其實就可以做答案了.. 要抽因子的話 [抽 (2/9)] = (2/9)[3ln(x)x^(3/2) - 2x^(3/2)] + C [再抽x^(3/2)] = (2/9)x^(3/2)[3ln(x) - 2] + C Note: 抽(2/9)時, (2/3)a = (2/9)(3a), (4/9)b = (2/9)(2b) 詳細: 要找 ∫x^(1/2)ln(x) dx 設 u = ln(x), du = 1/x dx 設 dv = x^(1/2) dx, v = [x^(3/2)]/(3/2) = (2/3)x^(3/2) 則 ∫x^(1/2)ln(x) dx = (2/3)ln(x)x^(3/2) - ∫[(2/3)x^(3/2)](1/x) dx = (2/3)ln(x)x^(3/2) - (2/3)∫x^(1/2) dx = (2/3)ln(x)x^(3/2) - (2/3){[x^(3/2)]/(3/2)} + C = (2/3)ln(x)x^(3/2) - (4/9)x^(3/2) + C = (2/9)[3ln(x)x^(3/2) - 2x^(3/2)] + C = (2/9)x^(3/2)[3ln(x) - 2] + C, 當中C是常數
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請教integral of sqrt(x)*log(x)發問:
integral of sqrt(x)*log(x) in到去(2/3)x^3/2 ln(x)-(4/9)(x^3/2)就唔知點做到去 (2/9)x^3/2 (3ln(x)-2) http://img40.imageshack.us/img40/6935/msp303591a493467i15adii.png
最佳解答:
= (2/3)ln(x)x^(3/2) - (4/9)x^(3/2) + C 其實就可以做答案了.. 要抽因子的話 [抽 (2/9)] = (2/9)[3ln(x)x^(3/2) - 2x^(3/2)] + C [再抽x^(3/2)] = (2/9)x^(3/2)[3ln(x) - 2] + C Note: 抽(2/9)時, (2/3)a = (2/9)(3a), (4/9)b = (2/9)(2b) 詳細: 要找 ∫x^(1/2)ln(x) dx 設 u = ln(x), du = 1/x dx 設 dv = x^(1/2) dx, v = [x^(3/2)]/(3/2) = (2/3)x^(3/2) 則 ∫x^(1/2)ln(x) dx = (2/3)ln(x)x^(3/2) - ∫[(2/3)x^(3/2)](1/x) dx = (2/3)ln(x)x^(3/2) - (2/3)∫x^(1/2) dx = (2/3)ln(x)x^(3/2) - (2/3){[x^(3/2)]/(3/2)} + C = (2/3)ln(x)x^(3/2) - (4/9)x^(3/2) + C = (2/9)[3ln(x)x^(3/2) - 2x^(3/2)] + C = (2/9)x^(3/2)[3ln(x) - 2] + C, 當中C是常數
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