高等数学，求不定积分。大神快来！！！

Answer 1

1、原式=∫4/(2sinxcosx*2cos^2x)dx
=∫4/[sin2x*(1+cos2x)]dx
令u=tanx，sin2x=2u/(1+u^2)，cos2x=(1-u^2)/(1+u^2)，dx=du/(1+u^2)
原式=∫4/[2u/(1+u^2)*2/(1+u^2)]*du/(1+u^2)
=∫(1+u^2)/udu
=∫(1/u+u)du
=ln|u|+u^2/2+C
=ln|tanx|+tan^2x/2+C，其中C是任意常数

2、原式=∫[(x+2)^2-4(x+2)+4]/(x+2)^3dx
=∫[1/(x+2)-4/(x+2)^2+4/(x+2)^3]d(x+2)
=ln|x+2|+4/(x+2)-2/(x+2)^2+C，其中C是任意常数

Answer 2

【俊狼猎英】团队为您解答~
1）变形
∫1/sin*cos^3dx
=∫sec^2*sec^2/tandx
=∫(tan^2+1)/tand(tanx)
=∫tanxdtanx+∫dtanx/tanx
=(tanx)^2/2+ln|tanx|+C
2）拆分分子
x^2=(x+2)^2-4(x+2)+4
原积分=∫1/(x+2)-4/(x+2)^2+4/(x+2)^3dx
=ln(x+2)+4/(x+2)-2/(x+2)^2+C
=ln(x+2)+(4x+6)/(x+2)^2+C

Answer 3

(1)
∫dx/[sinx.(cosx)^3]

let
y = siny
dy = cosy dx

∫dx/[sinx.(cosx)^3]
=∫ cosx dx/[sinx.(cosx)^4]
=∫dy/[y(1-y^2)^2]

let
1/[y(1-y^2)^2] ≡ A/y + (By+C)/(1-y^2) + (Dy+E)/(1-y^2)^2
=>
1 ≡ A(1-y^2)^2 + (By+C)y(1-y^2) + (Dy+E)y

y=0 => A =1

y=1
D+E = 1 (1)
y=-1
D-E =1 (2)

(1)+(2)
D =1/2

(1)-(2)
E=0

coef. of y^4
A-B =0
B = 1

coef. of y
C +E =0
C=0

1/[y(1-y^2)^2] ≡ 1/y + y/(1-y^2) + (1/2)y/(1-y^2)^2

∫dx/[sinx.(cosx)^3]
=∫dy/[y(1-y^2)^2]

=∫ {1/y + y/(1-y^2) + (1/2)y/(1-y^2)^2} dy
= ln|y| - (1/2)ln|1-y^2|- (1/4)[1/(1-y^2)] + C
= ln|sinx| - ln|cosx|- (1/2)[1/cosx] + C

(2)
x^2
= (x+2)^2 - 4x-4
= (x+2)^2 - 4(x+2) +4

∫x^2/(x+2)^3 dx
=∫[1/(x+2) - 4/(x+2)^2 + 4/(x+2)^3] dx
=ln|x+2| +4/(x+2) - 2/(x+2)^2 + C