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

May 14, 2016

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

#### Explanation:

Let us first find partial fractions of $\frac{x}{\left(x - 1\right) \left(x + 1\right)}$ and for this let

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

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

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

i.e. $A + B = 1$ and $A - B = 0$ i.e. $A = B = \frac{1}{2}$

Hence intx/((x-1)(x+1))dx=int[1/(2(x-1))+1/(2(x+1)]dx

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