# How do you use partial fraction decomposition to decompose the fraction to integrate x/((x+7)(x+8)(x+9))?

Mar 31, 2018

$\int \frac{x}{\left(x + 7\right) \left(x + 8\right) \left(x + 9\right)} \mathrm{dx} = \frac{7}{2} \ln | x + 7 | + 8 \ln | x + 8 | - \frac{9}{2} \ln | x + 9 | + C$

#### Explanation:

We can decompose the integrand as follows:

$\frac{x}{\left(x + 7\right) \left(x + 8\right) \left(x + 9\right)} = \frac{A}{x + 7} + \frac{B}{x + 8} + \frac{C}{x + 9}$

$\frac{x}{\left(x + 7\right) \left(x + 8\right) \left(x + 9\right)} = \frac{A \left(x + 8\right) \left(x + 9\right)}{\left(x + 7\right) \left(x + 8\right) \left(x + 9\right)} + \frac{\left(B\right) \left(x + 7\right) \left(x + 9\right)}{\left(x + 7\right) \left(x + 8\right) \left(x + 9\right)} + \frac{\left(C\right) \left(x + 7\right) \left(x + 8\right)}{\left(x + 7\right) \left(x + 8\right) \left(x + 9\right)}$

Set numerators equal:

$x = A \left(x + 8\right) \left(x + 9\right) + B \left(x + 7\right) \left(x + 9\right) + C \left(x + 7\right) \left(x + 8\right)$

We need to find $A , B , C .$ Fortunately, it looks like we can do this by choosing particular values of $x$ which will eliminate some terms on the right side.

$x = - 8 :$

$- 8 = B \left(- 1\right) \left(1\right) , B = 8$

$x = - 7 :$

$- 7 = A \left(- 1\right) \left(2\right) , A = \frac{7}{2}$

$x = - 9 :$

$- 9 = C \left(- 2\right) \left(- 1\right) , C = - \frac{9}{2}$

So, with the decomposed fraction, the integral becomes

$\frac{7}{2} \int \frac{\mathrm{dx}}{x + 7} + 8 \int \frac{\mathrm{dx}}{x + 8} - \frac{9}{2} \int \frac{\mathrm{dx}}{x + 9}$

Note that all constants were factored out.

These are all simple integrals.

In general, $\int \frac{\mathrm{dx}}{x \pm a} = \ln | x \pm a | + C$

Integrating yields

$\int \frac{x}{\left(x + 7\right) \left(x + 8\right) \left(x + 9\right)} \mathrm{dx} = \frac{7}{2} \ln | x + 7 | + 8 \ln | x + 8 | - \frac{9}{2} \ln | x + 9 | + C$