# How do you use partial fractions to find the integral int 1/(x^2-4)dx?

Jan 24, 2017

$\int \setminus \frac{1}{{x}^{2} - 4} \setminus \mathrm{dx} = \frac{1}{4} \ln A \left\mid \frac{x - 2}{x + 2} \right\mid$

#### Explanation:

We can factorise the denominator of the integrand as follows:

$\frac{1}{{x}^{2} - 4} = \frac{1}{\left(x + 2\right) \left(x - 2\right)}$

And we can decompose this into partial fractions of the form;

$\frac{1}{\left(x + 2\right) \left(x - 2\right)} \equiv \frac{A}{x + 2} + \frac{B}{x - 2}$
$\text{ } = \frac{A \left(x - 2\right) + B \left(x + 2\right)}{\left(x + 2\right) \left(x - 2\right)}$
$\therefore \text{ } 1 = A \left(x - 2\right) + B \left(x + 2\right)$

Put $x = - 2 \implies 1 = - 4 A \implies A = - \frac{1}{4}$
Put $x = 2 \setminus \setminus \setminus \setminus \setminus \implies 1 = 4 B \setminus \setminus \setminus \setminus \implies B = \frac{1}{4}$

Hence the partial fraction decomposition is:

$\frac{1}{\left(x + 2\right) \left(x - 2\right)} = \frac{\frac{1}{4}}{x - 2} - \frac{\frac{1}{4}}{x + 2}$

And so the integral can be written;

$\int \setminus \frac{1}{{x}^{2} - 4} \setminus \mathrm{dx} = \int \setminus \frac{\frac{1}{4}}{x - 2} - \frac{\frac{1}{4}}{x + 2} \setminus \mathrm{dx}$

Which we can now easily integrate to get;

$\int \setminus \frac{1}{{x}^{2} - 4} \setminus \mathrm{dx} = \frac{1}{4} \ln \left\mid x - 2 \right\mid - \frac{1}{4} \ln \left\mid x + 2 \right\mid + C$
$\text{ } = \frac{1}{4} \ln \left\mid x - 2 \right\mid - \frac{1}{4} \ln \left\mid x + 2 \right\mid + \frac{1}{4} \ln A$
$\text{ } = \frac{1}{4} \ln A \left\mid \frac{x - 2}{x + 2} \right\mid$