Question

不定积分怎么求

第三个和第四个

Answer 1

∫［sinx/（1＋sinx）］dx
＝∫［（1＋sinx－1）/（1＋sinx）］dx
＝∫dx－∫［1/（1＋sinx）］dx
＝x－∫｛（1－sinx）/［1－（sinx）^2］｝dx
＝x－∫［1/（cosx）^2］dx＋∫［sinx/（cosx）^2］dx
＝x－tanx－∫［1/（cosx）^2］d（cosx）
＝x－tanx＋1/cosx＋C
∫xarctan²x dx
=(1/2)∫arctan²x d(x²)
=(1/2)x²arctan²x - ∫x²arctanx/(1+x²) dx
=(1/2)x²arctan²x - ∫(x²+1-1)arctanx/(1+x²) dx
=(1/2)x²arctan²x - ∫arctanx dx + ∫arctanx/(1+x²) dx
中间那个积分用分部积分,第三个积分直接凑微分
=(1/2)x²arctan²x - xarctanx + ∫ x/(1+x²) dx + ∫arctanx d(arctanx)
=(1/2)x²arctan²x - xarctanx + (1/2)∫ 1/(1+x²) d(x²) + (1/2)arctan²x
=(1/2)x²arctan²x - xarctanx + (1/2)ln(1+x²) + (1/2)arctan²x +C

Answer 2

(3) 令 tan(x/2) = u, 则 sinx = 2u/(1+u^2), dx = 2du/(1+u^2)
I = ∫[4udu/(1+u^2)^2]/[1+2u/(1+u^2)]
= ∫4udu/[(1+u)^2 (1+u^2)] = 2∫[1/(1+u^2)-1/(1+u)^2]du
= 2arctanu + 2/(1+u) + C = 2arctan[tan(x/2)] + 2/[1+tan(x/2)] + C
(4) I = ∫x(arctanx)^2dx = (1/2)∫(arctanx)^2dx^2
= (1/2)(xarctanx)^2 - ∫x^2(arctanx)dx/(1+x^2)
= (1/2)(xarctanx)^2 - ∫arctanxdx + ∫(arctanx)dx/(1+x^2)
= (1/2)(xarctanx)^2 - xarctanx +∫[x/(1+x^2)]dx + ∫(arctanx)darctanx
= (1/2)(xarctanx)^2 - xarctanx +(1/2)ln(1+x^2) + (1/2)(arctanx)^2 + C