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

Dec 11, 2016

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

#### Explanation:

Let's do the `partial fraction decomposition

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

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

Therefore,

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

Let $x = 0$, $\implies$, $1 = - A$, $\implies$, $A = - 1$

Let $x = 1$, $\implies$, $1 = 2 C$, $\implies$, $C = \frac{1}{2}$

Let $x = - 1$, $\implies$, $1 = 2 B$, $\implies$, $B = \frac{1}{2}$

So,

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

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

=-ln(∣x∣)+1/2ln(∣x+1∣)+ln(∣x-1∣)+C