Question

∫(dx/x^4+3x^2+2)如何计算

Answer 1

let
1/(x^4+3x^2+2) ≡ (Ax+B)/(x^2+1) +(Cx+D)/(x^2+2)
=>
1 ≡ (Ax+B)(x^2+2) +(Cx+D)(x^2+1)
x=0, => 2B+D = 1 (1)
coef. of x^2 => B+D =0 (2)

(1)-(2)
B= 1
from (2)
D=-1
coef. of x^3 => A+C =0 (3)

coef. of x => 2A+C =0 (4)
=> A=C =0
1/(x^4+3x^2+2) ≡ 1/(x^2+1) -1/(x^2+2)
∫dx/(x^4+3x^2+2)
= ∫ [1/(x^2+1) -1/(x^2+2) ] dx
= arctanx - (1/√2).arctan(x/√2) + C

Answer 2

∫dx/(x^4+3x^2+2) = ∫dx/[(x^2+1)(x^2+2)]
= ∫[1/(1+x^2)-1/(2+x^2)]dx
= arctanx - (1/√2)arctan(x/√2) + C