∫ xe^x/(1+x)^2 dx

Question

如题所述

Answer 1

∫ [x e^x /(x + 1)²] dx
=∫ (x e^x) (x + 1)^(-2) dx
=x e^x = u → [(1)e^x + x e^x] dx = du → (e^x + x e^x) dx = du → e^x (1 + x) dx = du
(x + 1)^(-2) dx = dv → (x + 1)^(-2) d(x + 1) = dv → [(x + 1)^(-2+1)] /(-2+1) = v → [(x + 1)^(-1)] /(-1) = v → [- 1 /(x + 1)] = v
∫ u dv = v u - ∫ v du → ∫ (x e^x) (x + 1)^(-2) dx = [- 1 /(x + 1)] x e^x - ∫ [- 1 /(x + 1)] e^x (1 + x) dx
∫ (x e^x) (x + 1)^(-2) dx = [- x e^x /(x + 1)] + ∫ e^x dx
=[- x e^x /(x + 1)] + e^x + c
= e^x {[- x /(x + 1)] + 1} + c
=e^x [(- x + x + 1)/(x + 1)] + c
e^x [1 /(x + 1)] + c

∫ [x e^x /(x + 1)²] dx = [e^x /(x + 1)] + c

Answer 2

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Answer 3

可尝试用分部积分法