Question

这个怎么积分，怎么分解有理分式

Answer 1

∫ x/(1+x^3) dx
let
x/(1+x^3) ≡ A/(x+1) + (Bx+C)/(x^2-x+1)
=>
x ≡ A(x^2-x+1) + (Bx+C)(x+1)
x=-1, => A =-1/3
coef. of x^2,
A+B=0
B = 1/3
coef. of constant
A+C =0
C= 1/3
x/(1+x^3) ≡ -(1/3)[1/(x+1)] +(1/3) [(x+1)/(x^2-x+1)]
∫ x/(1+x^3) dx
=∫{ (1/3)[1/(x+1)] +(1/3) [(x+1)/(x^2-x+1)] } dx
=-(1/3)ln|x+1| +(1/3) ∫[(x+1)/(x^2-x+1)] dx
=(1/3)ln|x+1| +(1/6) ∫[(2x-1)/(x^2-x+1)] dx +(1/6)∫dx/(x^2-x+1)
=(1/3)ln|x+1| +(1/6)ln|x^2-x+1| -(√3/9)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)∫ du
=(2√3/3)u + C'
=(2√3/3)arctan[(2x-1)√3] + C'

Answer 2

凑微分
=1/2∫d(x^2+1)/(x^2+1)
=1/2ln(x^2+1)+C

追问

下面三次方啊