Question

高数不定积分怎么求

求详细步骤

Answer 1

(tanx)’=(1/cosx)^2利用凑微分法所以原积分等于∫1+tanxd(tanx )=tanx+(tanx)^2/2+cc为任意常数望采纳

Answer 2

令 √x = u， 则 x =u^2, dx = 2udu
I = 2∫arcsinudu/(1-u) = -2∫arcsinudln|1-u|
= -2ln|1-u|arcsinu +2∫ln|1-u|du/√(1-u^2)
题目拟似有误。
若题目是 ∫arcsin√x dx / [√x√(1-x)]
令 √x = u， 则 x =u^2, dx = 2udu
则 I = 2∫arcsinudu/√(1-u^2) = 2∫arcsinudarcsinu
= (arcsinu)^2 + C = (arcsin√x )^2 + C

Answer 3

想像去求一个在坐标轴的图形面积，套用公式

Answer 4

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