# How do you integrate int (x-9)/((x+8)(x-7)(x-5))  using partial fractions?

Feb 13, 2016

$- \frac{17}{195} \ln \left(x + 8\right) - \frac{1}{15} \ln \left(x - 7\right) + \frac{2}{13} \ln \left(x - 5\right) + {C}_{1}$

#### Explanation:

To Integrate, partial fractions are required, which can be done as follows: Let,
$\frac{x - 9}{\left(x + 8\right) \left(x - 7\right) \left(x - 5\right)} = \frac{A}{x + 8} + \frac{B}{x - 7} + \frac{C}{x - 5}$

=$\frac{A \left(x - 7\right) \left(x - 5\right) + B \left(x + 8\right) \left(x - 5\right) + C \left(x + 8\right) \left(x - 7\right)}{\left(x + 8\right) \left(x - 7\right) \left(x - 5\right)}$

Thus$x - 9 = A \left({x}^{2} - 12 x + 35\right) + B \left({x}^{2} + 3 x - 40\right) + C \left({x}^{2} + x - 56\right)$

Comparing like coefficients, it would be
A+B+C=0, -12A +3B +C=1 and 35A -40B-56C=-9. Solving for A,B and C, it would be

$A = - \frac{17}{195} , B = - \frac{1}{15} \mathmr{and} C = \frac{2}{13}$

The given integral would thus be $- \frac{17}{195} \ln \left(x + 8\right) - \frac{1}{15} \ln \left(x - 7\right) + \frac{2}{13} \ln \left(x - 5\right) + {C}_{1}$