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標題:
Integration Problem
Prove that ∫(e^x)/(x+1) dx=(e^x)/(x+1) + ∫(e^x)/(x+1)^2 dx. Hence find ∫x(e^x)/(x+1)^2 dx. Please help the second part. Can't think of a solution. Thank you.
最佳解答:
(a).Put u=1/(x+1), du=(-1)(x+1)^(-2) =-1/(x+1)2dxPut dv=(e^x)dx, v=e^xSo, sub. u, v, du, dv into ∫udv=uv-∫vdu ∫(e^x)/(x+1)dx=[1/(x+1)](e^x)-∫(e^x)[-1/(x+1)2dx]=(e^x)/(x+1)+∫(e^x)/(x+1)2dx (b).∫x(e^x)/(x+1)2dx=∫(x+1-1)(e^x)/(x+1)2dx=∫[(x+1)(e^x)/(x+1)2-(e^x)/(x+1)2]dx=∫[(e^x)/(x+1)-(e^x)/(x+1)2]dx=∫(e^x)/(x+1)dx-∫(e^x)/(x+1)2dx=(e^x)/(x+1)+∫(e^x)/(x+1)2dx-∫(e^x)/(x+1)2dx=(e^x)/(x+1)
其他解答:
Integration Problem
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發問:Prove that ∫(e^x)/(x+1) dx=(e^x)/(x+1) + ∫(e^x)/(x+1)^2 dx. Hence find ∫x(e^x)/(x+1)^2 dx. Please help the second part. Can't think of a solution. Thank you.
最佳解答:
(a).Put u=1/(x+1), du=(-1)(x+1)^(-2) =-1/(x+1)2dxPut dv=(e^x)dx, v=e^xSo, sub. u, v, du, dv into ∫udv=uv-∫vdu ∫(e^x)/(x+1)dx=[1/(x+1)](e^x)-∫(e^x)[-1/(x+1)2dx]=(e^x)/(x+1)+∫(e^x)/(x+1)2dx (b).∫x(e^x)/(x+1)2dx=∫(x+1-1)(e^x)/(x+1)2dx=∫[(x+1)(e^x)/(x+1)2-(e^x)/(x+1)2]dx=∫[(e^x)/(x+1)-(e^x)/(x+1)2]dx=∫(e^x)/(x+1)dx-∫(e^x)/(x+1)2dx=(e^x)/(x+1)+∫(e^x)/(x+1)2dx-∫(e^x)/(x+1)2dx=(e^x)/(x+1)
其他解答:
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