1/(2-x)(1-x)^1/2不定积分？

Answer 1

通过配项化简可得：
x/2-1/4+1/(4-8x);
之后逐项积分咯；
结果：x^2/4-x/4-1/8*ln(x-0.5)+C

Answer 2

令 u = x^ (1/2)
∫ (u²+1)^ (1/2) du , 令 u = tant, du = sec²t dt, (u²+1)^ (1/2) = sect
= ∫ sec³t dt
I = ∫ sect d(tant) = sect tant - ∫ tant * sect tant dt
= sect tant - ∫ tan²t * sect dt
= sect tant - ∫ (sec³t - sect) dt = sect tant - I + ∫ sect dt
2 I = sect tant + ln|sect + tant | + C
I = (1/2) [ sect tant + ln|sect + tant | } + C
= (1/2) [ u * (u²+1)^ (1/2) + ln | u + (u²+1)^ (1/2) |] + C
= (1/2) * √x * √(1+x) + (1/2) ln | √x + √(1+x) | + C

Answer 3

[(x+1)^1/2-(x-1)^1/2]/[(x+1)^1/2+(x-1)^1/2]=x - √(x^2-1)

积分得
x^2/2 - 1/2 x √[-1 + x^2] + 1/2 Log[x + √[-1 + x^2]]

Answer 4

∫√[(1-x)/（1+x)]dx
=∫(1-x)/√(1-x^2)dx
=∫1/√(1-x^2)-∫x/√(1-x^2)dx
=arcsinx+1/2∫(1-x^2)^(-1/2)d(1-x^2)
=arcsinx+√(1-x^2)+c

Answer 5

let
1/[x(1+x)(1+x+x^2)] ≡A/x+B/(x+1) +(Cx+D)/(x^2+x+1)
=>
1 ≡A(1+x)(1+x+x^2)+Bx(1+x+x^2) +(Cx+D)x(1+x)
x=0, => A = 1/3
x=-1, =>B=-1
coef. of x^3

A+B+C =0
1/3 -1 + C=0
C= -2/3
x=1
6A + 3B + 2(C+D) = 1
2-3 - 4/3 + 2D =1
D = 5/3
1/[x(1+x)(1+x+x^2)]

≡(1/3)(1/x)- 1/(x+1) + (1/3)[(-2x+5)/(x^2+x+1)]

∫dx/[x(1+x)(1+x+x^2)]
=∫{ (1/3)(1/x)- 1/(x+1) +(1/3) [(-2x+5)/(x^2+x+1) } dx
=(1/3)ln|x| - ln|x+1| +(1/3) ∫(-2x+5)/(x^2+x+1) dx
=(1/3)ln|x| - ln|x+1| -(1/3) ∫(2x+1)/(x^2+x+1) dx +2∫dx/(x^2+x+1) dx
=(1/3)ln|x| - ln|x+1| -(1/3)ln|x^2+x+1| +2∫dx/(x^2+x+1) dx

=(1/3)ln|x| - ln|x+1| -(1/3)ln|x^2+x+1| +(4√3/3)arctan[( 2x+1)/√3] + C
consider
x^2+x+1 = (x +1/2)^2 + 3/4
let
x+1/2 =(√3/2)tanu
dx =(√3/2)(secu)^2 du
∫dx/(x^2+x+1)
=∫(√3/2)(secu)^2 du/[ (3/4) (secu)^2 ]
= (2√3/3) u + C
= (2√3/3)arctan[( 2x+1)/√3] + C