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

Dec 9, 2016

The answer is ==-1/2ln(∣x+1∣)+1/2ln(∣x-1∣)+C

#### Explanation:

We use ${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$

The denominator is

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

So, the decompostion into partial fractions is

$\frac{1}{{x}^{2} - 1} = \frac{A}{x + 1} + \frac{B}{x - 1}$

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

Therefore,

$1 = A \left(x - 1\right) + B \left(x + 1\right)$

Let $x = 1$, $\implies$, $1 = 2 B$

Let $x = - 1$, $\implies$,$1 = - 2 A$

so,

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

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

=-1/2ln(∣x+1∣)+1/2ln(∣x-1∣)+C