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

Sep 21, 2017

$- \frac{1}{9} \ln | \left(x - 2\right) | + \frac{5}{18} \ln | \left(x + 7\right) | - \frac{1}{6} \ln | \left(x + 1\right) | + C$

#### Explanation:

$\int \frac{1 - 2 x}{\left(x - 2\right) \left(x + 7\right) \left(x + 1\right)} \mathrm{dx}$

consider the integrand. The denominators are all linear so we have a straightforward identity

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

multiply both sides by$\left(\left(x - 2\right) \left(x + 7\right) \left(x + 1\right)\right)$

after cancelling we end up with

$\left(1 - 2 x\right) \equiv A \left(x + 7\right) \left(x + 1\right) + B \left(x - 2\right) \left(x + 1\right) + C \left(x - 2\right) \left(x + 7\right)$

$\text{let } x = 2$

$1 - 4 = A \times 9 \times 3 + 0 + 0$

$- 3 = 27 A$

$\implies A = - \frac{1}{9}$

$\text{let } x = - 7$

$1 - \left(2 \times - 7\right) = 0 + B \times - 9 \times - 6 + 0$

$15 = 54 B$

$\implies B = \frac{15}{54} = \frac{5}{18}$

$\text{let } x = - 1$

$1 - \left(2 \times - 1\right) = 0 + 0 + C \times \left(- 1 - 2\right) \left(- 1 + 7\right)$

$\implies 3 = - 18 C$

$\implies C = - \frac{1}{6}$

$\therefore \int \frac{1 - 2 x}{\left(x - 2\right) \left(x + 7\right) \left(x + 1\right)} \mathrm{dx}$

$\int \left(- \frac{1}{9 \left(x - 2\right)} + \frac{5}{18 \left(x + 7\right)} - \frac{1}{6 \left(x + 1\right)}\right) \mathrm{dx}$

using$\textcolor{b l u e}{\int \frac{f ' \left(x\right)}{f \left(x\right)} \mathrm{dx} = \ln | f \left(x\right) | + C}$

we have

$- \frac{1}{9} \ln | \left(x - 2\right) | + \frac{5}{18} \ln | \left(x + 7\right) | - \frac{1}{6} \ln | \left(x + 1\right) | + C$