1/(1-x2) can be split into the partial fractions A/(1+x) + B/(1-x), where A and B are real constants, which when evaluated by multiplying the equation 1/(1-x2) = A/(1+x) + B/(1-x) through by (1-x2) = (1+x)(1-x) and substituting x =1, and x = -1; we find A = B = 0.5 hence 1/(1-x2) = 1/2(1-x) + 1/2(1+x) which can easily be integrated to 0.5( -log(1-x) + log(1+x)) + c or in the more accepted form 0.5(log(1+x) - log(1-x)) + c. (Where c is a real constant).
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