大神帮忙做做这道题 有理函数的不定积分

如图

1/[(x-1)(x^2+1)^2]≡ A/(x-1) +(Bx+C)/(x^2+1) + (Dx+E)/(x^2+1)^2
=>
1≡ A(x^2+1)^2 +(Bx+C)(x-1)(x^2+1) + (Dx+E)(x-1)
x=1, =>A=1/4
coef. of x^4,

A+B=0 =>B =-1/4
------------

coef. of x^3,
-B+C =0
C=B= -1/4
-----------

coef. of constant
A+C-E =1
1/4 -1/4 -E =1
E=-1
--------

x=-1
1≡ 4A -4(-B+C) - 2(-Dx+E)
1=1 -2(-D-1)
1= 1+2D+2
D= -1
1/[(x-1)(x^2+1)^2]
≡ (1/4)[1/(x-1)]-(1/4)[(x+1)/(x^2+1)] - [(x+1)/(x^2+1)^2]
∫dx/[(x-1)(x^2+1)^2]
=∫ {(1/4)[1/(x-1)]-(1/4)[(x+1)/(x^2+1)] - [(x+1)/(x^2+1)^2]} dx
=(1/4)ln|x-1| - (1/8)ln|x^2+1| - (1/4)arctanx + (1/2)[1/(x^2+1)] -∫ dx/(x^2+1)^2
=(1/4)ln|x-1| - (1/8)ln|x^2+1| - (1/4)arctanx + (1/2)[1/(x^2+1)]

-(1/2) [ arctanx + x/(x^2+1)] +C
=(1/4)ln|x-1| - (1/8)ln|x^2+1| - (5/8)arctanx + (1/2)[(1-x)/(x^2+1)] +C

let
x = tanu
dx = (secu)^2 du
∫ dx/(x^2+1)^2
=∫ (secu)^2 du/(secu)^4
=∫ (cosu)^2 du
=(1/2)∫ (1+cos2u) du
=(1/2) [ u + (1/2)sin2u] +C
=(1/2) [ arctanx + x/(x^2+1)] +C