# How do you integrate int (1-2x^2)/((x-2)(x+6)(x-7))  using partial fractions?

Dec 17, 2016

The answer is =7/40ln(∣x-2∣)-71/104ln(∣x+6∣)-97/65ln(∣x-7∣)+C

#### Explanation:

Let's do the decomposition into partial fractions

$\frac{1 - 2 {x}^{2}}{\left(x - 2\right) \left(x + 6\right) \left(x - 7\right)} = \frac{A}{x - 2} + \frac{B}{x + 6} + \frac{C}{x - 7}$

$= \frac{A \left(x + 6\right) \left(x - 7\right) + B \left(x - 2\right) \left(x - 7\right) + C \left(x - 2\right) \left(x + 6\right)}{\left(x - 2\right) \left(x + 6\right) \left(x - 7\right)}$

Therefore,

$1 - 2 {x}^{2} = A \left(x + 6\right) \left(x - 7\right) + B \left(x - 2\right) \left(x - 7\right) + C \left(x - 2\right) \left(x + 6\right)$

Let $x = 2$, $\implies$, $- 7 = - 40 A$, $\implies$, $A = \frac{7}{40}$

Let $x = - 6$, $\implies$, $- 71 = 104 B$, $\implies$. $B = - \frac{71}{104}$

Let $x = 7$, $\implies$, $- 97 = 65 C$, $\implies$, $C = - \frac{97}{65}$

So,

$\frac{1 - 2 {x}^{2}}{\left(x - 2\right) \left(x + 6\right) \left(x - 7\right)} = \frac{\frac{7}{40}}{x - 2} - \frac{\frac{71}{104}}{x + 6} - \frac{\frac{97}{65}}{x - 7}$

$\int \frac{\left(1 - 2 {x}^{2}\right) \mathrm{dx}}{\left(x - 2\right) \left(x + 6\right) \left(x - 7\right)} = \int \frac{\frac{7}{40} \mathrm{dx}}{x - 2} - \int \frac{\frac{71}{104} \mathrm{dx}}{x + 6} - \int \frac{\frac{97}{65} \mathrm{dx}}{x - 7}$

=7/40ln(∣x-2∣)-71/104ln(∣x+6∣)-97/65ln(∣x-7∣)+C