How do you integrate $int (1-x^2)/((x-9)(x-3)(x-2))$ using partial fractions?

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Answer 1 · 2016-11-02T19:45:55

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$(1 - x^2)/((x - 9)(x - 3)(x - 2)) = A/(x - 9) + B/(x - 3) + C/(x - 2)$

Multiply both sides by the common denominator:

$1 - x^2 = A(x - 3)(x - 2) + B(x - 9)(x - 2) + C(x - 9)(x - 3)$

Let $x = 9$ to make B and C disappear:

$1 - 9^2 = A(9 - 3)(9 - 2)$

$-80 = A(6)(7)$

$A = -40/21$

Let $x = 3$ to make A and C disappear:

$1 - 3^2 = B(3 - 9)(3 - 2)$

$-8 = B(-6)(1)$

$B = 4/3$

Let $x = 2$ to make A and B disappear:

$1 - 2^2 = C(2 - 9)(2 - 3)$

$-3 = C(-7)(-1)$

$C = -3/7$

$int(1 - x^2)/((x - 9)(x - 3)(x - 2))dx = -40/21int1/(x - 9) + 4/3int1/(x - 3) -3/7int1/(x - 2)$

$int(1 - x^2)/((x - 9)(x - 3)(x - 2))dx = -40/21ln|x - 9| + 4/3ln|x - 3| -3/7ln|x - 2| + C$

How do you integrate int (1-x^2)/((x-9)(x-3)(x-2)) int (1-x^2)/((x-9)(x-3)(x-2)) using partial fractions?