# How do you integrate #(x^2-1)/(x^4-16)# using partial fractions?

Partial Fractions. So, let,

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To integrate ( \frac{x^2 - 1}{x^4 - 16} ) using partial fractions, first factor the denominator ( x^4 - 16 ) as ( (x^2 - 4)(x^2 + 4) ). Then, express the fraction as a sum of partial fractions:

[ \frac{x^2 - 1}{x^4 - 16} = \frac{A}{x - 2} + \frac{B}{x + 2} + \frac{Cx + D}{x^2 + 4} ]

Solve for the constants ( A ), ( B ), ( C ), and ( D ) by equating coefficients. After finding the values of ( A ), ( B ), ( C ), and ( D ), integrate each term separately.

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