先令(1+x)/(1-x)=u,则变形后得x=(u-1)/(u+1),故dx=d(u-1)/(u+1)=2/[(u+1)^2]du原式中1/(1-x^2）则代入u，化为：(u+1)^2/4u故原式整体化为：∫2/[(u+1)^2]*(u+1)^2/4u*lnudu=∫1/(2u)lnudu=1/2∫lnudlnu=1/4*(lnu)^2+c=1/4*(ln(1+x)/(1-x))^2+c
