Question

请问微积分这题怎么做？

怎么继续算下去?

Answer 1

如下图所示，供参考，不用换成x，还得换回去

Answer 2

1/(x^3-x) = 1/[x(x-1)(x+1)]
let
1/[x(x-1)(x+1)]≡ A/x +B/(x-1) +C/(x+1)
=>
1≡ A(x-1)(x+1) +Bx(x+1) +Cx(x-1)
x=0, => A=-1
x=1, => B = 1/2
x=-1, => C=1/2

1/[x(x-1)(x+1)]≡ -1/x +(1/2)[1/(x-1)] +(1/2)[1/(x+1)]
∫dx/[x(x-1)(x+1)]
=∫ { -1/x +(1/2)[1/(x-1)] +(1/2)[1/(x+1)] } dx
=-ln|x| +(1/2)ln|x-1| + (1/2)ln|x+1| + C

Answer 3