∫(x^2+1)/[ (x-1)(1+x)^2]dx
let
(x^2+1)/[ (x-1)(1+x)^2]≡A/(x-1) +B/(1+x) +C/(1+x)^2
=>
x^2+1≡A(1+x)^2 +B(x-1)(1+x) +C(x-1)
x=1,
4A =2
A=1/2
x= -1
-2C = 2
C=-1
coef.of x^2
A+B= 1
B=1/2
=>
(x^2+1)/[ (x-1)(1+x)^2]≡(1/2)[1/(x-1)] +(1/2)[1/(x+1)] -1/(1+x)^2
∫(x^2+1)/[ (x-1)(1+x)^2]dx
=∫[(1/2)[1/(x-1)] +(1/2)[1/(x+1)] -1/(1+x)^2]dx
=(1/2(ln|(x-1)/(x+1) | + 1/(1+x) + C