1/1-x2 不定积分

∫ dx/(1-x²)
令1/(1-x²) = A/(1+x) + B/(1-x)
则1 = A(1-x) + B(1+x)
当x=-1,1=2A => A=1/2
当x=1,1=2B => B=1/2
∴原式= (1/2)∫ dx/(1+x) + ∫ dx/(1-x)
= (1/2)∫ d(1+x)/(1+x) ∫ -d(1-x)/(1-x)
= (1/2)ln|1+x| - (1/2)ln|1-x| + c
= (1/2)ln|(1+x)/(1-x)| + c