Question

求 ∫1/[(x^2+1)*(x^2+x+1)] dx

Answer 1

let
1/[(x^2+1)*(x^2+x+1)] = (Ax+B)/(x^2+1) + (Cx+D)/(x^2+x+1)
=>1=(Ax+B)(x^2+x+1) + (Cx+D)(x^2+1)
put x=0
B+D=1 (1)
coef.of x^3
A+C=0 (2)
coef.of x^2
A+B+D=0 (3)
(1)+(2)
A+B+C+D=1 (4)
from (3) and (4)
C=1
from (2) => A=-1
coef.of x
A+B+C=0
=>B=0
from (1) => D=1
1/[(x^2+1)*(x^2+x+1)] = -x/(x^2+1) + (x+1)/(x^2+x+1)
∫1/[(x^2+1)*(x^2+x+1)] dx
= ∫[-x/(x^2+1) + (x+1)/(x^2+x+1)] dx
=∫ (-1/2) d(x^2+1)/(x^2+1) + (1/2) ∫ (2x+1)/(x^2+x+1)] dx + (1/2)∫ (1/(x^2+x+1))dx
= (-1/2)ln(x^2+1) + (1/2) ln(x^2+x+1) + (1/2)∫ (1/(x^2+x+1))dx
=(-1/2)ln(x^2+1) + (1/2) ln(x^2+x+1) + (1/2)∫ (1/[(x+1/2)^2+3/4])dx
let
x+1/2 = (√3/2) tany
dx = (√3/2) (secy)^2dy
∫ (1/[(x+1/2)^2+3/4])dx
= (4/3)∫ [1/(secy)^2][(√3/2) (secy)^2dy]
=(2√3/3) y
=(2√3/3) arctan[2√3(x+1/2)/3]
therefore
∫1/[(x^2+1)*(x^2+x+1)] dx
=(-1/2)ln(x^2+1) + (1/2) ln(x^2+x+1) + (1/2)∫ (1/[(x+1/2)^2+3/4])dx
=(-1/2)ln(x^2+1) + (1/2) ln(x^2+x+1) + (√3/3) arctan[2√3(x+1/2)/3] + C