∫dx/(1-x^2)=(1/2)∫[1/(1-x)+1/(1+x)]dx=(1/2)∫dx/(1-x)+(1/2)∫dx/(x+1)=-(1/2)∫d(-x+1)/(-x+1)+(1/2)∫dx/(x+1)=-(1/2)ln|1-x|+(1/2)ln|1+x|+c=(1/2)ln|(1+x)/(1-x)|+c原理：∫dx/x=lnx+c.