高等数学 求解 不定积分？

Question

如题所述

Answer 1

详细过程如图所示………

Answer 2

∫ [(x+2)/(x-1)]^2 . (1/x) dx
=∫ [1 + 3/(x-1)]^2 . (1/x) dx
=∫ [1 + 6/(x-1) + 9/(x-1)^2] . (1/x) dx
=∫ { 1/x + 6/[x(x-1)] + 9/[x(x-1)^2] } dx
=∫ { 1/x + 6[1/(x-1) - 1/x] + 9[ 1/x -1/(x-1) + 1/(x-1)^2 ] } dx
=∫ [ 4/x -3/(x-1) +9/(x-1)^2 ] dx
=4ln|x| -3ln|x-1| -9/(x-1) + C
//

let

1/[x(x-1)^2]≡ A/x +B/(x-1) + C/(x-1)^2
=>
1≡ A(x-1)^2 +Bx(x-1) + Cx
x=0, => A = 1
x=1, => C=1
coef. of x^2
A+B=0
B=-1
1/[x(x-1)^2]
≡ A/x +B/(x-1) + C/(x-1)^2
≡ 1/x -1/(x-1) + 1/(x-1)^2本回答被网友采纳

Answer 3

方法如下图所示：追答