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

Oct 28, 2016

The integral is $= \frac{1}{4} \left(\ln \left(x - 2\right) - \ln \left(x + 2\right)\right) + C$

#### Explanation:

Let factorise the denominator
${x}^{2} - 4 = \left(x - 2\right) \left(x + 2\right)$
So we have
$\frac{1}{{x}^{2} - 4} = \frac{1}{\left(x - 2\right) \left(x + 2\right)}$
So the decomposition into partial fractions is
$\frac{1}{{x}^{2} - 4} = \frac{1}{\left(x - 2\right) \left(x + 2\right)} = \frac{A}{x - 2} + \frac{B}{x + 2}$

=(A(x+2)+B(x-2))/((x-2)(x+2)

So equalising LHS and RHS
$1 = A \left(x + 2\right) + B \left(x - 2\right)$

If $x = 2$ then $1 = 4 A$$\implies$$A = \frac{1}{4}$
and $x = - 2$ then $1 = - 4 B$$\implies$ $B = - \frac{1}{4}$

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

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

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