Partial Fractions

A rational integrand with a factorable quadratic denominator calls for partial fraction decomposition, the technique of splitting one fraction into simpler reciprocal pieces matched to each denominator factor. Begin by factoring x^2 - 1 = (x-1)(x+1) and writing \frac{1}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1}. To find the constants, clear denominators to get 1 = A(x+1) + B(x-1): substituting the root x = 1 isolates A = 1/2, while substituting x = -1 isolates B = -1/2. The integral then separates into \frac{1}{2}\int\frac{dx}{x-1} - \frac{1}{2}\int\frac{dx}{x+1} = \frac{1}{2}(\ln|x-1| - \ln|x+1|) + C = \frac{1}{2}\ln\left|\frac{x-1}{x+1}\right| + C. The option with the reciprocal argument flips the sign of the constant difference. Option \ln|x^2-1| omits the crucial factor of 1/2 and the correct logarithmic ratio. Option \arctan x applies instead to 1/(x^2+1), the opposite-sign quadratic that does not factor over the reals. This decomposition is the canonical rational-function integration pattern. As a final plausibility check, differentiating the answer term by term recovers \frac{1}{2}\left(\frac{1}{x-1} - \frac{1}{x+1}\right) = \frac{1}{x^2-1}, which matches the original integrand exactly and is valid everywhere except at the excluded points |x| = 1 where the denominator vanishes.