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

Nov 2, 2016

#### Explanation:

Expand

$\frac{1 - {x}^{2}}{\left(x - 9\right) \left(x - 3\right) \left(x - 2\right)} = \frac{A}{x - 9} + \frac{B}{x - 3} + \frac{C}{x - 2}$

Multiply both sides by the common denominator:

$1 - {x}^{2} = A \left(x - 3\right) \left(x - 2\right) + B \left(x - 9\right) \left(x - 2\right) + C \left(x - 9\right) \left(x - 3\right)$

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

$1 - {9}^{2} = A \left(9 - 3\right) \left(9 - 2\right)$

$- 80 = A \left(6\right) \left(7\right)$

$A = - \frac{40}{21}$

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

$1 - {3}^{2} = B \left(3 - 9\right) \left(3 - 2\right)$

$- 8 = B \left(- 6\right) \left(1\right)$

$B = \frac{4}{3}$

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

$1 - {2}^{2} = C \left(2 - 9\right) \left(2 - 3\right)$

$- 3 = C \left(- 7\right) \left(- 1\right)$

$C = - \frac{3}{7}$

$\int \frac{1 - {x}^{2}}{\left(x - 9\right) \left(x - 3\right) \left(x - 2\right)} \mathrm{dx} = - \frac{40}{21} \int \frac{1}{x - 9} + \frac{4}{3} \int \frac{1}{x - 3} - \frac{3}{7} \int \frac{1}{x - 2}$

$\int \frac{1 - {x}^{2}}{\left(x - 9\right) \left(x - 3\right) \left(x - 2\right)} \mathrm{dx} = - \frac{40}{21} \ln | x - 9 | + \frac{4}{3} \ln | x - 3 | - \frac{3}{7} \ln | x - 2 | + C$