Question

求不定积分∫dx/(x^2+1)(x^2+x+1)

Answer 1

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