Integer of (1+x²)/abs(1-x²)

If 1 - x² => 0 :

int ( 1 + x² ) / | 1 - x² | dx

= int ( 1 + x² ) / ( 1 - x² ) dx

= int ( x² - 1 + 2 ) / ( 1 - x² ) dx

= int -1 + 2 / ( 1 - x² ) dx

= int -1 + 2 / [(1 - x)(1 + x)] dx

2 / [(1 - x)(1 + x)] = a / (1 - x) + b / (1 + x)

=> 2 = a(1 + x) + b(1 - x) = a + ax + b - bx = (a - b)x + a + b

=> a = b and a + b = 2

=> a = 1 and b = 1

So :

int ( 1 + x² ) / | 1 - x² | dx

= int -1 + 1 / (1 - x) + 1 / (1 + x) dx

= int -1 - (-1) / (1 - x) + 1 / (1 + x) dx

= [-x - ln(|1 - x|) + ln(|1 + x|)]

Can you do 1 - x² < 0 ?