# How do you integrate #int (3x^3+2x^2-7x-6)/(x^2-4) dx# using partial fractions?

The integral is

Start by using long division to divide. We want the degree of the numerator to be *less* than the degree of the denominator.

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To integrate ( \frac{3x^3 + 2x^2 - 7x - 6}{x^2 - 4} ) using partial fractions, follow these steps:

- Factor the denominator ( x^2 - 4 ) as ( (x - 2)(x + 2) ).
- Write the given fraction as a sum of partial fractions with undetermined constants ( A ) and ( B ):

[ \frac{3x^3 + 2x^2 - 7x - 6}{x^2 - 4} = \frac{A}{x - 2} + \frac{B}{x + 2} ]

- Multiply both sides by the denominator ( x^2 - 4 ) to clear the fractions.
- Solve for ( A ) and ( B ) by equating coefficients of like terms.
- Once you find ( A ) and ( B ), integrate each partial fraction separately.
- Finally, combine the integrals to get the result.

The steps above will yield the integral of ( \frac{3x^3 + 2x^2 - 7x - 6}{x^2 - 4} ) using partial fractions.

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