# How do you find int x/((x+2)(x+9))dx using partial fractions?

##### 1 Answer
Nov 3, 2015

$\frac{9}{7} \ln \left(x + 9\right) - \frac{2}{7} \ln \left(x + 2\right) +$constant

#### Explanation:

The equation to be integrated can be written as:

x/((x+2)(x+9))=A/(x+2)+B/(x+9)=(A(x+9)+B(x+2))/((x+2)(x+9)

Equating coefficients

$x \to A \left(x + 9\right) + B \left(x + 2\right) = \left(A + B\right) x + \left(9 A + 2 B\right)$

$x = \left(A + B\right) x$

Therefore, dividing by $x$:

• $A + B = 1$ $\left(1\right)$

Additionally,

• $9 A + 2 B = 0$ $\left(2\right)$

Simultaneous Equations

$\left(2\right) - 2 \times \left(1\right)$

$9 A + 2 B = 0$ $\left(2\right)$
$- 2 A - 2 B = - 2$ $\left(1\right)$

Cancel out the $B$ terms:

$7 A = - 2$

$A = - \frac{2}{7}$

Subbing $A$ back into $\left(1\right)$:

$B = \frac{9}{7}$

Integrating

Subbing $A$ and $B$ back into the original equation:

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

$= \int \frac{9}{7 \left(x + 9\right)} \mathrm{dx} - \int \frac{2}{7 \left(x + 2\right)} \mathrm{dx}$

$= \frac{9}{7} \int \frac{1}{x + 9} \mathrm{dx} - \frac{2}{7} \int \frac{1}{x + 2} \mathrm{dx}$

$= \frac{9}{7} \ln \left(x + 9\right) +$constant $- \frac{2}{7} \ln \left(x + 2\right) +$constant

$= \frac{9}{7} \ln \left(x + 9\right) - \frac{2}{7} \ln \left(x + 2\right) +$constant