# How do you integrate #int (1-x^2)/((x-9)(x-5)(x+2))dx # using partial fractions?

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To integrate ( \int \frac{1-x^2}{(x-9)(x-5)(x+2)} ) using partial fractions, we first perform polynomial long division to ensure that the degree of the numerator is less than the degree of the denominator. After that, we express the rational function as a sum of partial fractions. Once we've found the appropriate form of the partial fraction decomposition, we can solve for the unknown coefficients by equating numerators and comparing coefficients. Finally, we integrate each term separately.

Since the degree of the numerator is less than the degree of the denominator, we can proceed with the partial fraction decomposition.

Let's express ( \frac{1-x^2}{(x-9)(x-5)(x+2)} ) as:

[ \frac{1-x^2}{(x-9)(x-5)(x+2)} = \frac{A}{x-9} + \frac{B}{x-5} + \frac{C}{x+2} ]

Multiplying both sides by the denominator, we have:

[ 1-x^2 = A(x-5)(x+2) + B(x-9)(x+2) + C(x-9)(x-5) ]

We'll now solve for ( A ), ( B ), and ( C ) by substituting appropriate values of ( x ) that make two of the terms zero.

After finding the values of ( A ), ( B ), and ( C ), we integrate each term separately, and then combine them to obtain the final result.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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