Question

一道高数题目

Answer 1

如图所示

追问

为啥要乘6啊😁

追答

复合函数

请及时采纳一下我的回答

谢谢

追问

好滴，谢谢

Answer 2

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