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

Nov 29, 2016

$\int \frac{x}{\left(x + 3\right) \left(x + 6\right) \left(x + 2\right)} \mathrm{dx} = \ln | x + 3 | - \frac{1}{2} \ln | x + 6 | - \frac{1}{2} \ln | x + 2 | + C$

#### Explanation:

The Partial Fraction decomposition of the integrand will be of the form:

$\frac{x}{\left(x + 3\right) \left(x + 6\right) \left(x + 2\right)} \equiv \frac{A}{x + 3} + \frac{B}{x + 6} + \frac{C}{x + 2}$
$\therefore \frac{x}{\left(x + 3\right) \left(x + 6\right) \left(x + 2\right)} = \frac{A \left(x + 6\right) \left(x + 2\right) + B \left(x + 3\right) \left(x + 2\right) + C \left(x + 3\right) \left(x + 6\right)}{\left(x + 3\right) \left(x + 6\right) \left(x + 2\right)}$
$\therefore x = A \left(x + 6\right) \left(x + 2\right) + B \left(x + 3\right) \left(x + 2\right) + C \left(x + 3\right) \left(x + 6\right)$

Put $\left\{\begin{matrix}x = - 3 & \implies - 3 = A \left(3\right) \left(- 1\right) & \implies A = 1 \\ x = - 6 & \implies - 6 = B \left(- 3\right) \left(- 4\right) & \implies B = - \frac{1}{2} \\ x = - 2 & \implies - 2 = C \left(1\right) \left(4\right) & \implies C = - \frac{1}{2}\end{matrix}\right.$

Hence,

$\frac{x}{\left(x + 3\right) \left(x + 6\right) \left(x + 2\right)} \equiv \frac{1}{x + 3} - \frac{1}{2 \left(x + 6\right)} - \frac{1}{2 \left(x + 2\right)}$

So,

$\int \frac{x}{\left(x + 3\right) \left(x + 6\right) \left(x + 2\right)} \mathrm{dx} = \int \frac{1}{x + 3} - \frac{1}{2 \left(x + 6\right)} - \frac{1}{2 \left(x + 2\right)} \mathrm{dx}$
$\therefore \int \frac{x}{\left(x + 3\right) \left(x + 6\right) \left(x + 2\right)} \mathrm{dx} = \ln | x + 3 | - \frac{1}{2} \ln | x + 6 | - \frac{1}{2} \ln | x + 2 | + C$