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

Jul 18, 2016

$\ln | x + 1 | + \ln | x - 1 | + C$where C is a constant

#### Explanation:

The given expression can be written as partial sum of fractions:

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

Now let's integrate :

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

$\int \frac{1}{x + 1} + \frac{1}{x - 1} \mathrm{dx}$

$\int \frac{1}{x + 1} \mathrm{dx} + \int \frac{1}{x - 1} \mathrm{dx}$

$\int \frac{d \left(x + 1\right)}{x + 1} + \int \frac{d \left(x - 1\right)}{x - 1}$

$\ln | x + 1 | + \ln | x - 1 | + C$where C is a constant