Question

1/(1-X^2)的积分

Answer 1

∫dx/(1-x^2)
=∫dx/(1+x)(1-x)
=∫dx(1/(1+x)+1/(1-x)
=∫dx/(1+x)+∫dx/(1-x)
=∫d(x+1)/(1+x)-∫d(x-1)/(x-1)
=ln(x+1)-ln(x-1)+C
=ln[(x+1)/(x-1)]+C