Question

不定积分2x-5/(x-1)²(x+2)？

Answer 1

2x-5=3(x-1)-(x+2)
拆开积分即可
∫(2x-5)/[(x-1)²(x+2)]dx
=∫3/[(x-1)(x+2)]dx-∫1/(x-1)²dx
=∫[(x+2)-(x-1)]/[(x-1)(x+2)]dx-∫(x-1)^{-2}dx
=∫1/(x-1)dx-∫1/(x+2)dx-∫(x-1)^{-2}d(x-1)
=ln|x-1|-ln|x+2}+1/(x-1)+C

Answer 2

(2x-5)/[(x-1)^2.(x+2)]
let
(2x-5)/[(x-1)^2.(x+2)] ≡ A/(x-1) +B/(x-1)^2 +C/(x+2)
=>
2x-5 ≡ A(x-1)(x+2) +B(x+2) +C(x-1)^2
x=-2, =>C=-1
x=1, =>B=-1
coef. of x^2
A+C=0
A=1
ie
(2x-5)/[(x-1)^2.(x+2)] ≡ 1/(x-1) -1/(x-1)^2 -1/(x+2)
∫ (2x-5)/[(x-1)^2.(x+2)] dx
=∫ [ 1/(x-1) -1/(x-1)^2 -1/(x+2) ] dx
=ln|x-1| + 1/(x-1) -ln|x+2| +C