Integrate(1+x)/((1-x^2)(2x+1)) with respect to x.
To simplify the fraction first notice (1-x2) = (1-x)(1+x) so the common factor of (1+x) in the numerator and denominator can be cancelled. (1+x)/((1-x^2)(2x+1)) = 1/((1-x)(2x+1)), then we need to split this fractions into partials. 1/((1-x)(2x+1)) = A/(1-x) + B/(2x+1) which implies 1 = A(2x+1) + B(1-x). Setting x = 1, 1 = 3A so A = 1/3. Setting x = -1/2, 1 = 3B/2 so B = 2/3. So we must integrate (1/3(1-x) + 2/3(2x+1)), the integral of 2/3(2x+1) is (ln|2x+1|)/3 since the numerator is one third of the derivative of the denominator. Similarly the integral of 1/3(1-x) is (-ln|1-x|)/3. So the answer is (ln|2x+1| - ln|1-x|)/3 + c
RH
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