# How do you use partial fractions to find the integral #int 1/(x^2-1)dx#?

The answer is

The denominator is

So, the decompostion into partial fractions is

Therefore,

so,

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To integrate 1/(x^2 - 1) dx using partial fractions, first decompose the fraction into its partial fractions. Rewrite 1/(x^2 - 1) as A/(x - 1) + B/(x + 1). Then, find the values of A and B by equating numerators: A(x + 1) + B(x - 1) = 1. Solve for A and B. Once you have A and B, integrate each partial fraction separately. The integral of A/(x - 1) is A ln |x - 1|, and the integral of B/(x + 1) is B ln |x + 1|. Therefore, the integral of 1/(x^2 - 1) dx is A ln |x - 1| + B ln |x + 1| + C, where C is the constant of integration.

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