Question

求这道不定积分问题

求这道不定积分问题求给个解题步骤

Answer 1

x^3+1 = (x+1)(x^2-x+1)
3/(x^3+1) ≡ A/(x+1) +(Bx+C)/(x^2-x+1)
=>
3≡ A(x^2-x+1) +(Bx+C)(x+1)
x=-1 => A=1
coef. of x^2
A+B=0
B=-1
coef. of constant
A+C =3
C=2
3/(x^3+1) ≡ 1/(x+1) +(-x+2)/(x^2-x+1)
∫ 3/(x^3+1) dx
=∫ [ 1/(x+1) +(-x+2)/(x^2-x+1) ] dx
=ln|x+1| -∫ (x-2)/(x^2-x+1) ] dx
=ln|x+1| -(1/2)∫ (2x-1)/(x^2-x+1) dx +(3/2)∫ dx/(x^2-x+1)
=ln|x+1| -(1/2)ln|x^2-x+1| +(3/2)∫ dx/(x^2-x+1)
=ln|x+1| -(1/2)ln|x^2-x+1| +√3 .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

怎么看不见
我的答案怎么样？

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