# How do you integrate int (x-3x^2)/((x-6)(x-2)(x+4))  using partial fractions?

Jan 9, 2017

The answer is =-51/20ln(∣x-6∣)+5/12ln(∣x-2∣)-13/15ln(∣x+4∣)+C

#### Explanation:

Let's perform the decomposition into partial fractions

$\frac{x - 3 {x}^{2}}{\left(x - 6\right) \left(x - 2\right) \left(x + 4\right)} = \frac{A}{x - 6} + \frac{B}{x - 2} + \frac{C}{x + 4}$

$= \frac{A \left(x - 2\right) \left(x + 4\right) + B \left(x - 6\right) \left(x + 4\right) + C \left(x - 6\right) \left(x - 2\right)}{\left(x - 6\right) \left(x - 2\right) \left(x + 4\right)}$

Therefore,

$x - 3 {x}^{2} = A \left(x - 2\right) \left(x + 4\right) + B \left(x - 6\right) \left(x + 4\right) + C \left(x - 6\right) \left(x - 2\right)$

Let $x = 6$, $\implies$, $- 102 = 40 A$, $\implies$, $A = - \frac{51}{20}$

Let $x = 2$, $\implies$, $- 10 = - 24 B$, $\implies$, $B = \frac{5}{12}$

Let $x = - 4$, $\implies$, $- 52 = 60 C$, $\implies$, $C = - \frac{13}{15}$

So,

$\frac{x - 3 {x}^{2}}{\left(x - 6\right) \left(x - 2\right) \left(x + 4\right)} = \frac{- \frac{51}{20}}{x - 6} + \frac{\frac{5}{12}}{x - 2} + \frac{- \frac{13}{15}}{x + 4}$

Therefore,

$\int \frac{\left(x - 3 {x}^{2}\right) \mathrm{dx}}{\left(x - 6\right) \left(x - 2\right) \left(x + 4\right)} = - \frac{51}{20} \int \frac{\mathrm{dx}}{x - 6} + \frac{5}{12} \int \frac{\mathrm{dx}}{x - 2} - \frac{13}{15} \int \frac{\mathrm{dx}}{x + 4}$

=-51/20ln(∣x-6∣)+5/12ln(∣x-2∣)-13/15ln(∣x+4∣)+C