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

Nov 14, 2017

$\frac{31}{49} \cdot L n \left(x - 2\right) - \frac{80}{49} \cdot \ln \left(x - 9\right) - \frac{3}{7} \cdot {\left(x - 2\right)}^{- 1} + C$

#### Explanation:

I partitioned integrand into basic fractions,

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

After expanding denominator,

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

After setting $x = 2$, $- 7 C = - 3$, so $C = \frac{3}{7}$

After setting $x = 9$, $49 A = - 80$, so $A = - \frac{80}{49}$

After setting $x = 1$, $A + 8 B - 8 C = 0$, so $B = \frac{1}{8} \cdot \left(8 C - A\right) = \frac{31}{49}$

Thus,

$\int \frac{1 - {x}^{2}}{\left(x - 9\right) \cdot {\left(x - 2\right)}^{2}} \cdot \mathrm{dx}$

=$- \frac{80}{49} \cdot \int \frac{\mathrm{dx}}{x - 9} + \frac{31}{49} \cdot \int \frac{\mathrm{dx}}{x - 2} + \frac{3}{7} \cdot \int \frac{\mathrm{dx}}{x - 2} ^ 2$

=$\frac{31}{49} \cdot L n \left(x - 2\right) - \frac{80}{49} \cdot \ln \left(x - 9\right) - \frac{3}{7} \cdot {\left(x - 2\right)}^{- 1} + C$