# How do you integrate int(x+1)/((x-9)(x+3)(x-2)) using partial fractions?

Apr 26, 2016

Set up a partial fraction decomposition and choose $x$-values to find the numerators of the partial fractions to get $\int \frac{x + 1}{\left(x - 9\right) \left(x + 3\right) \left(x - 2\right)} \mathrm{dx} = \frac{5}{42} \ln \left\mid x - 9 \right\mid - \frac{1}{30} \ln \left\mid x + 3 \right\mid - \frac{3}{35} \ln \left\mid x - 2 \right\mid + C$.

#### Explanation:

Since all the factors in the denominator of $\frac{x + 1}{\left(x - 9\right) \left(x + 3\right) \left(x - 2\right)}$ are linear (yay!), our partial fraction decomposition will be:
$\frac{A}{x - 9} + \frac{B}{x + 3} + \frac{C}{x - 2}$

For now, let's ignore the integral and focus on decomposing this fraction:
$\frac{x + 1}{\left(x - 9\right) \left(x + 3\right) \left(x - 2\right)} = \frac{A}{x - 9} + \frac{B}{x + 3} + \frac{C}{x - 2}$

$x + 1 = A \left(x + 3\right) \left(x - 2\right) + B \left(x - 9\right) \left(x - 2\right) + C \left(x - 9\right) \left(x + 3\right)$

Let $x = 9$ to find the value of $A$:
$9 + 1 = A \left(9 + 3\right) \left(9 - 2\right) + B \left(9 - 9\right) \left(9 - 2\right) + C \left(9 - 9\right) \left(9 + 3\right)$
$10 = 84 A$
$A = \frac{10}{84} = \frac{5}{42}$

Let $x = - 3$ to find the value of $B$:
$- 3 + 1 = A \left(- 3 + 3\right) \left(- 3 - 2\right) + B \left(- 3 - 9\right) \left(- 3 - 2\right) + C \left(- 3 - 9\right) \left(- 3 + 3\right)$
$- 2 = 60 B$
$B = - \frac{2}{60} = - \frac{1}{30}$

And let $x = 2$ to find the value of $C$:
$2 + 1 = A \left(2 + 3\right) \left(2 - 2\right) + B \left(2 - 9\right) \left(2 - 2\right) + C \left(2 - 9\right) \left(2 + 3\right)$
$3 = - 35 C$
$C = - \frac{3}{35}$

Therefore, $\frac{x + 1}{\left(x - 9\right) \left(x + 3\right) \left(x - 2\right)} = \frac{\frac{5}{42}}{x - 9} - \frac{\frac{1}{30}}{x + 3} - \frac{\frac{3}{35}}{x - 2}$. Our integral is now:
$\int \frac{\frac{5}{42}}{x - 9} - \frac{\frac{1}{30}}{x + 3} - \frac{\frac{3}{35}}{x - 2} \mathrm{dx}$

Using the sum rule, this becomes:
$\int \frac{\frac{5}{42}}{x - 9} \mathrm{dx} - \int \frac{\frac{1}{30}}{x + 3} \mathrm{dx} - \int \frac{\frac{3}{35}}{x - 2} \mathrm{dx}$

Integrating these gives:
$\frac{5}{42} \int \frac{1}{x - 9} \mathrm{dx} - \frac{1}{30} \int \frac{1}{x + 3} \mathrm{dx} - \frac{3}{35} \int \frac{1}{x - 2} \mathrm{dx}$
$\frac{5}{42} \ln \left\mid x - 9 \right\mid - \frac{1}{30} \ln \left\mid x + 3 \right\mid - \frac{3}{35} \ln \left\mid x - 2 \right\mid + C$