# How do you integrate int (1-x^2)/((x+8)(x-5)(x+2))  using partial fractions?

Nov 4, 2016

The answer is $= - \frac{21}{26} \ln \left(x + 8\right) - \frac{24}{91} \ln \left(x - 5\right) + \frac{1}{14} \ln \left(x + 2\right) + C$

#### Explanation:

Let's start by the decomposition into partial fractions
$\frac{1 - {x}^{2}}{\left(x + 8\right) \left(x - 5\right) \left(x + 2\right)} = \frac{A}{x + 8} + \frac{B}{x - 5} + \frac{C}{x + 2}$
Upon simplification
$1 - {x}^{2} = A \left(x - 5\right) \left(x + 2\right) + B \left(x + 8\right) \left(x + 2\right) + C \left(x + 8\right) \left(x - 5\right)$
Let $x = 5$$\implies$$- 24 = 91 B$$\implies$$B = - \frac{24}{91}$
Let $x = - 2$$\implies$$- 3 = - 42 C$$\implies$$C = \frac{1}{14}$
Let $x = - 8$$\implies$$- 63 = 78 A$$\implies$$A = - \frac{21}{26}$
$\int \frac{\left(1 - {x}^{2}\right) \mathrm{dx}}{\left(x + 8\right) \left(x - 5\right) \left(x + 2\right)} = \int \frac{- \frac{21}{26} \mathrm{dx}}{x + 8} + \int \frac{- \frac{24}{91} \mathrm{dx}}{x - 5} + \int \frac{\frac{1}{14} \mathrm{dx}}{x + 2}$
$= - \frac{21}{26} \ln \left(x + 8\right) - \frac{24}{91} \ln \left(x - 5\right) + \frac{1}{14} \ln \left(x + 2\right) + C$