# How do you integrate int(x+1)/((x-9)(x+8)(x-2)) using partial fractions?

##### 1 Answer
Jul 10, 2017

The answer is $= \frac{10}{119} \ln \left(| x - 9 |\right) - \frac{7}{170} \ln \left(| x + 8 |\right) - \frac{3}{70} \ln \left(| x - 2 |\right) + C$

#### Explanation:

Let's perform the decomposition into partial fractions

$\frac{x + 1}{\left(x - 9\right) \left(x + 8\right) \left(x - 2\right)} = \frac{A}{x - 9} + \frac{B}{x + 8} + \frac{C}{x - 2}$

$= \frac{A \left(x - 2\right) \left(x + 8\right) + B \left(x - 9\right) \left(x - 2\right) + C \left(x - 9\right) \left(x + 8\right)}{\left(x - 9\right) \left(x + 8\right) \left(x - 2\right)}$

The denominator is the same, we compare the numerator.

$x + 1 = A \left(x - 2\right) \left(x + 8\right) + B \left(x - 9\right) \left(x - 2\right) + C \left(x - 9\right) \left(x + 8\right)$

Let $x = 9$, $\implies$, $10 = 17 \cdot 7 A$, $\implies$, $A = \frac{10}{119}$

Let $x = - 8$, $\implies$, $- 7 = - 17 \cdot - 10 B$, $\implies$, $B = - \frac{7}{170}$

Let $x = 2$, $\implies$, $3 = - 7 \cdot 10 C$, $\implies$, $C = - \frac{3}{70}$

Therefore,

$\frac{x + 1}{\left(x - 9\right) \left(x + 8\right) \left(x - 2\right)} = \frac{\frac{10}{119}}{x - 9} - \frac{\frac{7}{170}}{x + 8} - \frac{\frac{3}{70}}{x - 2}$

$\int \frac{\left(x + 1\right) \mathrm{dx}}{\left(x - 9\right) \left(x + 8\right) \left(x - 2\right)} = \int \frac{\frac{10}{119} \mathrm{dx}}{x - 9} - \int \frac{\frac{7}{170} \mathrm{dx}}{x + 8} - \int \frac{\frac{3}{70} \mathrm{dx}}{x - 2}$

$= \frac{10}{119} \ln \left(| x - 9 |\right) - \frac{7}{170} \ln \left(| x + 8 |\right) - \frac{3}{70} \ln \left(| x - 2 |\right) + C$