# How do you integrate int (1-x^2)/((x-9)(x+6)(x-4))  using partial fractions?

Sep 10, 2016

$\int \frac{1 - {x}^{2}}{\left(x - 9\right) \left(x + 6\right) \left(x - 4\right)} \mathrm{dx} =$

$= - \frac{16}{15} \ln \left\mid x - 9 \right\mid - \frac{7}{30} \ln \left\mid x + 6 \right\mid + \frac{3}{10} \ln \left\mid x - 4 \right\mid + C$

#### Explanation:

$\frac{1 - {x}^{2}}{\left(x - 9\right) \left(x + 6\right) \left(x - 4\right)} = \frac{A}{x - 9} + \frac{B}{x + 6} + \frac{C}{x - 4}$

We can determine $A$, $B$ and $C$ using Heaviside's cover up method:

$A = \frac{1 - {\textcolor{b l u e}{9}}^{2}}{\left(\textcolor{b l u e}{9} + 6\right) \left(\textcolor{b l u e}{9} - 4\right)} = \frac{- 80}{\left(15\right) \left(5\right)} = - \frac{16}{15}$

$B = \frac{1 - {\textcolor{b l u e}{\left(- 6\right)}}^{2}}{\left(\textcolor{b l u e}{\left(- 6\right)} - 9\right) \left(\textcolor{b l u e}{\left(- 6\right)} - 4\right)} = \frac{- 35}{\left(- 15\right) \left(- 10\right)} = - \frac{7}{30}$

$C = \frac{1 - {\textcolor{b l u e}{4}}^{2}}{\left(\textcolor{b l u e}{4} - 9\right) \left(\textcolor{b l u e}{4} + 6\right)} = \frac{- 15}{\left(- 5\right) \left(10\right)} = \frac{3}{10}$

So:

$\int \frac{1 - {x}^{2}}{\left(x - 9\right) \left(x + 6\right) \left(x - 4\right)} \mathrm{dx} =$

$\int - \frac{16}{15} \left(\frac{1}{x - 9}\right) - \frac{7}{30} \left(\frac{1}{x + 6}\right) + \frac{3}{10} \left(\frac{1}{x - 4}\right) \mathrm{dx}$

$= - \frac{16}{15} \ln \left\mid x - 9 \right\mid - \frac{7}{30} \ln \left\mid x + 6 \right\mid + \frac{3}{10} \ln \left\mid x - 4 \right\mid + C$

where $C$ is the constant of integeration (not to be confused with the coefficient $C$ we calculated earlier).