Question

求不定积分。 然后用待定系数法。可是我不会写成这种形式。还有为什么这样分解啊 有什么方法 能教我下

求不定积分。 然后用待定系数法。可是我不会写成这种形式。还有为什么这样分解啊 有什么方法 能教我下吗

Answer 1

f(x) =x^3-3x+2
f(1)=0
x^3-3x+2 = (x-1)(x^2+ax-2)
coef. of x
-2-a=-3
a=1

x^3-3x+2 = (x-1)(x^2+x-2)=(x-1)^2.(x+2)

x/(x^3-3x+2) =x/[(x-1)^2.(x+2)]
let
x/[(x-1)^2.(x+2)]≡ A/(x-1) + B/(x-1)^2 + C/(x+2)
=>
x≡ A(x-1)(x+2) + B(x+2) + C(x-1)^2

x=1, =>B=1/3
x=-2, =>C = -2/9

coef. of x^2
A+C= 0
A=2/9

x/[(x-1)^2.(x+2)]≡ (2/9)[1/(x-1)] +(1/3) [1/(x-1)^2] -(2/9) [1/(x+2)]

∫x/(x^3-3x+2) dx
=(2/9)∫[1/(x-1)] dx +(1/3) ∫[1/(x-1)^2] dx -(2/9)∫ [1/(x+2)] dx
= (2/9)ln|(x-1)/(x+2)| -(1/3)[1/(x-1)] + C

Answer 2

你前面错了吧

追答

化简之后是这样的