# How do you integrate int (x-3x^2)/((x+9)(x-5)(x-7))  using partial fractions?

Oct 24, 2017

$\ln \left(\frac{{\left(x + 9\right)}^{\frac{9}{8}} {\left(x - 5\right)}^{\frac{5}{2}}}{x - 7} ^ \left(\frac{7}{4}\right)\right) + C$

#### Explanation:

First you'll need to split up the integrand into partial fractions.

$\frac{x - 3 {x}^{2}}{\left(x + 9\right) \left(x - 5\right) \left(x - 7\right)} = \frac{A}{x + 9} + \frac{B}{x - 5} + \frac{C}{x - 7}$

$x - 3 {x}^{2} = A \left(x - 5\right) \left(x - 7\right) + B \left(x + 9\right) \left(x - 7\right) + C \left(x + 9\right) \left(x - 5\right)$

We'll substitute in the roots, $x = - 9 , 5 , 7$, to find each of the numerator constants.

$x = - 9$

$\left(- 9\right) - 3 \times {\left(- 9\right)}^{2} = A \times \left(- 9 - 5\right) \times \left(- 9 - 7\right)$

$- 252 = A \times - 14 \times - 16$

$A = \frac{9}{8}$

$x = 5$

$\left(5\right) - 3 {\left(5\right)}^{2} = B \times \left(5 + 9\right) \times \left(5 - 7\right)$

$- 70 = B \times 14 \times - 2$

$B = \frac{5}{2}$

$x = 7$

$\left(7\right) - 3 {\left(7\right)}^{2} = C \times \left(7 + 9\right) \times \left(7 - 2\right)$

$- 140 = C \times 16 \times 5$

$C = - \frac{7}{4}$

Therefore, we now have that

$\frac{x - 3 {x}^{2}}{\left(x + 9\right) \left(x - 5\right) \left(x - 7\right)} = \frac{\frac{9}{8}}{x + 9} + \frac{\frac{5}{2}}{x - 5} - \frac{\frac{7}{4}}{x - 7}$

Integrating $\frac{a}{x + b}$ gives you $a \ln \left(x + b\right)$, so

$\int \frac{\frac{9}{8}}{x + 9} + \frac{\frac{5}{2}}{x - 5} - \frac{\frac{7}{4}}{x - 7} \mathrm{dx} =$

$\frac{9}{8} \ln \left(x + 9\right) + \frac{5}{2} \ln \left(x - 5\right) - \frac{7}{4} \ln \left(x - 7\right) + C$

$= \ln \left(\frac{{\left(x + 9\right)}^{\frac{9}{8}} {\left(x - 5\right)}^{\frac{5}{2}}}{x - 7} ^ \left(\frac{7}{4}\right)\right) + C$