∫ [1/(1-x^2)] * ln[(1+x)/(1-x)] dx
=(1/2)∫ [1/(1+x) + 1/(1-x)] * ln[(1+x)/(1-x)] dx
=(1/2)∫ ln[(1+x)/(1-x)] d[ln(1+x)-ln(1-x)]
=(1/2)∫ ln[(1+x)/(1-x)] d{ln[(1+x)/(1-x)]}
=(1/4) {ln[(1+x)/(1-x)]}^2 +C