# How do you integrate int (1-x^2)/((x-9)(x-5)(x+12))  using partial fractions?

Sep 12, 2016

$\int \frac{1 - {x}^{2}}{\left(x - 9\right) \left(x - 5\right) \left(x + 12\right)} \mathrm{dx}$

$= - \frac{20}{21} \ln \left\mid x - 9 \right\mid + \frac{6}{17} \ln \left\mid x - 5 \right\mid - \frac{143}{357} \ln \left\mid x + 12 \right\mid + \text{constant}$

#### Explanation:

Let's look at the general problem:

$\int \frac{a {x}^{2} + b x + c}{\left(x - d\right) \left(x - e\right) \left(x - f\right)} \mathrm{dx}$

where $d , e , f$ are distinct.

We can split the integrand into partial fractions like this:

$\frac{a {x}^{2} + b x + c}{\left(x - d\right) \left(x - e\right) \left(x - f\right)} = \frac{A}{x - d} + \frac{B}{x - e} + \frac{C}{x - f}$

where $A , B , C$ can be determined using Heaviside's cover up method:

$A = \frac{a {\textcolor{b l u e}{d}}^{2} + b \textcolor{b l u e}{d} + c}{\left(\textcolor{b l u e}{d} - e\right) \left(\textcolor{b l u e}{d} - f\right)}$

$B = \frac{a {\textcolor{b l u e}{e}}^{2} + b \textcolor{b l u e}{e} + c}{\left(\textcolor{b l u e}{e} - d\right) \left(\textcolor{b l u e}{e} - f\right)}$

$C = \frac{a {\textcolor{b l u e}{f}}^{2} + b \textcolor{b l u e}{f} + c}{\left(\textcolor{b l u e}{f} - d\right) \left(\textcolor{b l u e}{f} - e\right)}$

Then:

$\int \frac{a {x}^{2} + b x + c}{\left(x - d\right) \left(x - e\right) \left(x - f\right)} \mathrm{dx}$

$= \int \left(\frac{A}{x - d} + \frac{B}{x - e} + \frac{C}{x - f}\right) \mathrm{dx}$

$= A \ln \left\mid x - d \right\mid + B \ln \left\mid x - e \right\mid + C \ln \left\mid x - f \right\mid + \text{constant}$

In our example:

$a = - 1$, $b = 0$, $c = 1$, $d = 9$, $e = 5$, $f = - 12$

So:

$A = \frac{- 80}{\left(4\right) \left(21\right)} = - \frac{20}{21}$

$B = \frac{- 24}{\left(- 4\right) \left(17\right)} = \frac{6}{17}$

$C = \frac{- 143}{\left(- 21\right) \left(- 17\right)} = - \frac{143}{357}$

So

$\int \frac{1 - {x}^{2}}{\left(x - 9\right) \left(x - 5\right) \left(x + 12\right)} \mathrm{dx}$

$= - \frac{20}{21} \ln \left\mid x - 9 \right\mid + \frac{6}{17} \ln \left\mid x - 5 \right\mid - \frac{143}{357} \ln \left\mid x + 12 \right\mid + \text{constant}$