# How do you integrate int (1-x^2)/((x-9)(x-5)(x+2))dx  using partial fractions?

Mar 5, 2016

$- \frac{20}{11} \ln | x - 9 | + \frac{6}{7} \ln | x - 5 | - \frac{3}{77} \ln | x + 2 | + c o n s t .$

#### Explanation:

We can rewrite the integrand expression in this way
$\frac{1 - {x}^{2}}{\left(x - 9\right) \left(x - 5\right) \left(x + 2\right)} = \frac{A}{x - 9} + \frac{B}{x - 5} + \frac{C}{x + 2}$

Making $x = 0 , 1 \mathmr{and} - 1$ (it's convenient to make $x = 1$ and $x = - 1$ since then $1 - {x}^{2} = 0$), we get

$\frac{1}{90} = \frac{A}{-} 9 + \frac{B}{-} 5 + \frac{C}{2}$
$0 = \frac{A}{-} 8 + \frac{B}{-} 4 + \frac{C}{3}$
$0 = \frac{A}{-} 10 + \frac{B}{-} 6 + \frac{C}{1}$

Or
$\left[\begin{matrix}- \frac{1}{9} & - \frac{1}{5} & \frac{1}{2} \\ - \frac{1}{8} & - \frac{1}{4} & \frac{1}{3} \\ - \frac{1}{10} & - \frac{1}{6} & 1\end{matrix}\right] \left[\begin{matrix}A \\ B \\ C\end{matrix}\right] = \left[\begin{matrix}\frac{1}{90} \\ 0 \\ 0\end{matrix}\right]$

Solving the system of variables
$\Delta = \left[\begin{matrix}- \frac{1}{9} & - \frac{1}{5} & \frac{1}{2} \\ - \frac{1}{8} & - \frac{1}{4} & \frac{1}{3} \\ - \frac{1}{10} & - \frac{1}{6} & 1\end{matrix}\right] = \frac{1}{36} + \frac{1}{150} + \frac{1}{96} - \left(\frac{1}{80} + \frac{1}{162} + \frac{1}{40}\right) = \frac{1800 + 432 + 675 - 810 - 400 - 1620}{64800} = \frac{77}{64800}$
$\Delta A = \left[\begin{matrix}\frac{1}{90} & - \frac{1}{5} & \frac{1}{2} \\ 0 & - \frac{1}{4} & \frac{1}{3} \\ 0 & - \frac{1}{6} & 1\end{matrix}\right] = - \frac{1}{360} - \left(- \frac{1}{1620}\right) = \frac{- 9 + 2}{3240} = - \frac{7}{3240}$
$\Delta B = \left[\begin{matrix}- \frac{1}{9} & \frac{1}{90} & \frac{1}{2} \\ - \frac{1}{8} & 0 & \frac{1}{3} \\ - \frac{1}{10} & 0 & 1\end{matrix}\right] = - \frac{1}{2700} - \left(- \frac{1}{720}\right) = \frac{- 4 + 15}{10800} = \frac{11}{10800}$
$\Delta C = \left[\begin{matrix}- \frac{1}{9} & - \frac{1}{5} & \frac{1}{90} \\ - \frac{1}{8} & - \frac{1}{4} & 0 \\ - \frac{1}{10} & - \frac{1}{6} & 0\end{matrix}\right] = \frac{1}{4320} - \left(\frac{1}{3600}\right) = \frac{5 - 6}{21600} = - \frac{1}{21600}$

So

$A = \frac{\Delta A}{\Delta} = \frac{- \frac{7}{3240}}{\frac{77}{64800}} = - \frac{1}{11} \cdot 20 = - \frac{20}{11}$
$B = \frac{\Delta B}{\Delta} = \frac{\frac{11}{10800}}{\frac{77}{64800}} = \frac{1}{7} \cdot 6 = \frac{6}{7}$
$C = \frac{\Delta C}{\Delta} = \frac{- \frac{1}{21600}}{\frac{77}{64800}} = - \frac{1}{77} \cdot 3 = - \frac{3}{77}$

Then the original expression becomes

$= - \frac{20}{11} \int \frac{\mathrm{dx}}{x - 9} + \frac{6}{7} \int \frac{\mathrm{dx}}{x - 5} - \frac{3}{77} \int \frac{\mathrm{dx}}{x + 2}$
$= - \frac{20}{11} \ln | x - 9 | + \frac{6}{7} \ln | x - 5 | - \frac{3}{77} \ln | x + 2 | + c o n s t .$