Question

求不定积分，因式分解怎么做，求过程。谢谢。

Answer 1

追答

追问

我记得有一种求极限的方法，直接能出来。

想不起来了

追答

你记错了，没有这种方法

追问

麻烦你了

这样做对不对

Answer 2

令A/(x+2)+ B/(x+1)+C/(x-1)=1/[(x+2)(x²-1)]
A(x+1)(x-1)+B((x+2)(x-1)+C(x+2)(x+1)=1
整理，得
A(-1)+B(-2)+C(+2)-1=0
(A+B+C)x²+(B+2C)x+(2C-A-2B-1)=0
令A+B+C=0
B+2C=0
2C-A-2B-1=0
解得A=1/5，B=-2/5，C=1/5
1/[(x+2)(x²-1)]
=(1/5)/(x+2) -(2/5)/(x+1)+ (1/5)/(x-1)
=(1/5)[1/(x-1)+1/(x+2) -2/(x+1)]
∫1/[(x+2)(x²-1)]dx
=(1/5)∫[1/(x-1)+1/(x+2) -2/(x+1)]dx
=(1/5)∫[1/(x-1)]d(x-1) +(1/5)∫[1/(x+2)]d(x+2) -(2/5)∫[1/(x+1)]d(x+1)
=(1/5)ln|x-1|+(1/5)ln|x+2|-(2/5)ln|x+1| +C
=(1/5)ln|(x-1)(x+2)/(x+1)²| +C

Answer 3

待定系数法
A十B十C=0
B十3C=0
令C=1，
则B=-3，A=2
此时，通分分子
常数项
-A-2B+2C=6。
还需要乘以1/6。

Answer 4

1/[(x+2)(x^2-1)]=1/[(x+2)(x-1)(x+1)]
let
1/[(x+2)(x-1)(x+1)]≡ A/(x+2) + B/(x-1)+C/(x+1)
1≡ A(x-1)(x+1) + B(x+2)(x+1)+C(x+2)(x-1)
x=-2
1= A(-2-1)(-2+1)
A=1/3
x=1
1=B(1+2)(1+1)
B =1/6
x=-1
1=C(-1+2)(-1-1)
C = -1/2
∫dx/[(x+2)(x^2-1)]
=∫[ (1/3)(1/(x+2)) + (1/6)(1/(x-1))-(1/2)(1/(x+1)) ]dx
=(1/3)ln|x+2| +(1/6)ln|x-1| -(1/2)ln|x+1| + C