求助关于广义积分（反常积分）敛散性的判断问题

令I=∫(0→2)1/(1-x^2)dx,x=1为奇点；考察积分I1=∫(0→1-△)1/(1-x^2)dx和I2=∫(1+△→2)1/(1-x^2)dx,△＞0为无穷小量；设I(△)=I1+I2,I1=(1/2)ln|(2-△)/△|,I2=(1/2)ln3-(1/2)ln|(2+△)/△|,所以I(△)=I1+I2=(1/2)ln3+(1/2)ln|(2-△)/(2+△)|,lim(△→0)I(△)=(1/2)ln3,所以I=lim(△→0)I(△),即I=(1/2)ln3.