(x乘以e的x次幂)/(x+1)的平方的不定积分

Question

如题所述

Answer 1

∫ xe^x/(x + 1)² dx
= ∫ [(x + 1) - 1]e^x/(x + 1)² dx
= ∫ e^x/(x + 1) dx - ∫ e^x/(x + 1)² dx
= ∫ e^x/(x + 1) dx - ∫ e^x d[- 1/(x + 1)]
= ∫ e^x/(x + 1) dx + e^x/(x + 1) - ∫ 1/(x + 1) d(e^x)，分部积分法
= ∫ e^x/(x + 1) dx + e^x/(x + 1) - ∫ e^x/(x + 1) dx
= e^x/(x + 1) + C本回答被提问者采纳

Answer 2

＝∫[xe^x/(x+1)²]dx=∫[e^x/(x+1)]dx-∫[e^x/(x+1)²]dx=∫[e^x/(x+1)]dx+e^x/(x+1)-∫[e^x/(x+1)]dx=e^x/(x+1)+C