# How do you integrate int (x-3x^2)/((x-7)(x-5)(x+4))  using partial fractions?

Sep 4, 2016

$\int \frac{x - 3 {x}^{2}}{\left(x - 7\right) \left(x - 5\right) \left(x + 4\right)} \mathrm{dx}$

$= - \frac{70}{11} \ln \left\mid x - 7 \right\mid + \frac{35}{9} \ln \left\mid x - 5 \right\mid - \frac{52}{99} \ln \left\mid x + 4 \right\mid + C$

#### Explanation:

$\frac{x - 3 {x}^{2}}{\left(x - 7\right) \left(x - 5\right) \left(x + 4\right)} = \frac{A}{x - 7} + \frac{B}{x - 5} + \frac{C}{x + 4}$

Using Heaviside's cover up method, we find:

$A = \frac{\left(7\right) - 3 {\left(7\right)}^{2}}{\left(\left(7\right) - 5\right) \left(\left(7\right) + 4\right)} = \frac{7 - 147}{2 \cdot 11} = - \frac{140}{22} = - \frac{70}{11}$

$B = \frac{\left(5\right) - 3 {\left(5\right)}^{2}}{\left(\left(5\right) - 7\right) \left(\left(5\right) + 4\right)} = \frac{5 - 75}{\left(- 2\right) \cdot 9} = \frac{- 70}{- 18} = \frac{35}{9}$

$C = \frac{\left(- 4\right) - 3 {\left(- 4\right)}^{2}}{\left(\left(- 4\right) - 7\right) \left(\left(- 4\right) - 5\right)} = \frac{- 4 - 48}{\left(- 11\right) \left(- 9\right)} = - \frac{52}{99}$

So:

$\int \frac{x - 3 {x}^{2}}{\left(x - 7\right) \left(x - 5\right) \left(x + 4\right)} \mathrm{dx}$

$= \int - \frac{70}{11 \left(x - 7\right)} + \frac{35}{9 \left(x - 5\right)} - \frac{52}{99 \left(x + 4\right)} \mathrm{dx}$

$= - \frac{70}{11} \ln \left\mid x - 7 \right\mid + \frac{35}{9} \ln \left\mid x - 5 \right\mid - \frac{52}{99} \ln \left\mid x + 4 \right\mid + C$

(Note: The $C$ here stands for the integration constant, not the coefficient we calculated).