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integration
發問:
我想問點用integration by part計 ∫x/(4+x^2)dx? 唔該:) 更新: ∫(1/2 arctan(x/2)) (1) dx 下一步點解?唔明...
最佳解答:
留上用的方法只是simple substitution, 並不是 by part, 要小心~ Integration by part: ∫x/(4+x^2)dx Theorem: ∫u dv/dx dx = uv - ∫v du/dx dx ...(*) Let u = x Let dv/dx = 1/(4+x^2) du/dx = dx/dx = 1 v = ∫dv/dx dx = ∫1/(4+x^2)dx=1/2 arctan(x/2) Substituting back into (*) ∫u dv/dx dx = (x) (1/2 arctan(x/2)) - ∫(1/2 arctan(x/2)) (1) dx = x/2(arctan(x/2)) - 1/2 (x*arctan(x/2)-ln(x^2+4)) = x/2(arctan(x/2)) - x/2(arctan(x/2)) + 1/2(ln(x^2+4)) = 1/2(ln(x^2+4)) 2011-04-24 21:06:02 補充： 基本上 1/2 ∫(arctan(x/2)) (1) dx = 1/2 (x*arctan(x/2)-ln(x^2+4)) 但詳盡解釋這個要比較長時間+用圖, 看你這只是5分題, 你還是自己看看數學書學一下如何Integrate arctan 這種function吧。 2011-04-25 18:36:47 補充： 我覺得是因為問者在做integration by part 的exercise吧? 1+1 = 2 , 2*1 也是=2, 如果你不用一下其他方法, 如何學新東西呢~
一定要用integration by part? 用integration by substitution似乎快d!|||||∫x(4+x^2)^-1 dx let u=4+x^2 =1/2 ∫ u^-1 du du=2x dx =1/2 ln(4+x^2) +C
integration
發問:
我想問點用integration by part計 ∫x/(4+x^2)dx? 唔該:) 更新: ∫(1/2 arctan(x/2)) (1) dx 下一步點解?唔明...
最佳解答:
留上用的方法只是simple substitution, 並不是 by part, 要小心~ Integration by part: ∫x/(4+x^2)dx Theorem: ∫u dv/dx dx = uv - ∫v du/dx dx ...(*) Let u = x Let dv/dx = 1/(4+x^2) du/dx = dx/dx = 1 v = ∫dv/dx dx = ∫1/(4+x^2)dx=1/2 arctan(x/2) Substituting back into (*) ∫u dv/dx dx = (x) (1/2 arctan(x/2)) - ∫(1/2 arctan(x/2)) (1) dx = x/2(arctan(x/2)) - 1/2 (x*arctan(x/2)-ln(x^2+4)) = x/2(arctan(x/2)) - x/2(arctan(x/2)) + 1/2(ln(x^2+4)) = 1/2(ln(x^2+4)) 2011-04-24 21:06:02 補充： 基本上 1/2 ∫(arctan(x/2)) (1) dx = 1/2 (x*arctan(x/2)-ln(x^2+4)) 但詳盡解釋這個要比較長時間+用圖, 看你這只是5分題, 你還是自己看看數學書學一下如何Integrate arctan 這種function吧。 2011-04-25 18:36:47 補充： 我覺得是因為問者在做integration by part 的exercise吧? 1+1 = 2 , 2*1 也是=2, 如果你不用一下其他方法, 如何學新東西呢~
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其他解答:一定要用integration by part? 用integration by substitution似乎快d!|||||∫x(4+x^2)^-1 dx let u=4+x^2 =1/2 ∫ u^-1 du du=2x dx =1/2 ln(4+x^2) +C
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