Question

求不定积分∫dx/(x^2+1)(x^2+x+1)

速度……急用

Answer 1

原式=∫(-x/(x^2+1)+(x+1)/(x^2+x+1))dx=-∫x/(x^2+1)dx+∫(x+1)/(x^2+x+1/4+3/4)dx=-1/2∫d(x^2+1)/(x^2+1)+∫(x+1)/((x+1/2)^2+3/4)dx=-1/2ln(x^2+1)+4/3∫(x+1)/(((2x+1)/√3)^2+1)=-1/2ln(x^2+1)+∫(2x/√3+1/√3)/((2x/√3+1/√3)^2+1)d(2x/√3+1/√3)+1/√3∫1/((2x/√3+1/√3)^2+1)d(2x/√3+1/√3)=-1/2ln(x^2+1)+1/2ln((2x/√3+1/√3)^2+1)+1/√3arctan(2x/√3+1/√3)+C=-1/2ln(x^2+1)+1/2ln(4/3)+1/2ln(x^2+x+1)+1/√3arctan(2x/√3+1/√3)+C=-1/2ln(x^2+1)+1/2ln(x^2+x+1)+1/√3arctan(2x/√3+1/√3)+C

Answer 2

1/[(x^2+1)(x^2+x+1)] = (a1x+a0)/(x^2+1) + (b1x+b0)/(x^2+x+1)
1=(a1x+a0)(x^2+x+1)+(b1x+b0)(x^2+1)
coef. of x^3
a1+b1=0 (1)
put x=0
a0+b0=1 (2)
coef. of x
a0+a1+b1= 0
a0=0 (a1+b1=0)
coef. of x^2
a0+a1+b0=0
a1=-1 (a0+b0=1)
b0=1
b1=1
1/[(x^2+1)(x^2+x+1)] = -x/(x^2+1) + (x+1)/(x^2+x+1)
∫dx/[(x^2+1)(x^2+x+1)]
=∫ [-x/(x^2+1) + (x+1)/(x^2+x+1)] dx
= -(1/2) ln(x^2+1) + (1/2)ln(x^2+x+1) + (1/2)∫ dx/(x^2+x+1)

x^2+x+1 = (x+1/2)^2 + 3/4
let
x+1/2 =(√3/2)tana
dx = (√3/2) (seca)^2 da
∫ dx/(x^2+x+1)
= 2√3/3 ∫ da
=2√3/3 arctan(x+1/2) + C'

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